Codeforces Round #478 (Div. 2) E. Hag's Khashba

E. Hag's Khashba

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Hag is a very talented person. He has always had an artist inside him but his father forced him to study mechanical engineering.

Yesterday he spent all of his time cutting a giant piece of wood trying to make it look like a goose. Anyway, his dad found out that he was doing arts rather than studying mechanics and other boring subjects. He confronted Hag with the fact that he is a spoiled son that does not care about his future, and if he continues to do arts he will cut his 25 Lira monthly allowance.

Hag is trying to prove to his dad that the wooden piece is a project for mechanics subject. He also told his dad that the wooden piece is a strictly convex polygon with $\mathrm{nn vertices.}$

Hag brought two pins and pinned the polygon with them in the $\mathrm{11-st\; and 22-nd\; vertices\; to\; the\; wall.\; His\; dad\; has qq queries\; to\; Hag\; of\; two\; types.}$

• $\mathrm{11 ff tt:\; pull\; a\; pin\; from\; the\; vertex ff,\; wait\; for\; the\; wooden\; polygon\; to\; rotate\; under\; the\; gravity\; force\; (if\; it\; will\; rotate)\; and\; stabilize.\; And\; then\; put\; the\; pin\; in\; vertex tt.}$
• $\mathrm{22 vv:\; answer\; what\; are\; the\; coordinates\; of\; the\; vertex vv.}$

Please help Hag to answer his father's queries.

You can assume that the wood that forms the polygon has uniform density and the polygon has a positive thickness, same in all points. After every query of the 1-st type Hag's dad tries to move the polygon a bit and watches it stabilize again.

Input

The first line contains two integers $\mathrm{nn and qq (3\le n\le 100003\le n\le 10000, 1\le q\le 2000001\le q\le 200000) —\; the\; number\; of\; vertices\; in\; the\; polygon\; and\; the\; number\; of\; queries.}$

The next $\mathrm{nn lines\; describe\; the\; wooden\; polygon,\; the ii-th\; line\; contains\; two\; integers xixi and yiyi (|xi|,|yi|\le 108|xi|,|yi|\le 108) —\; the\; coordinates\; of\; the ii-th\; vertex\; of\; the\; polygon.\; It\; is\; guaranteed\; that\; polygon\; is\; strictly\; convex\; and\; the\; vertices\; are\; given\; in\; the\; counter-clockwise\; order\; and\; all\; vertices\; are\; distinct.}$

The next $\mathrm{qq lines\; describe\; the\; queries,\; one\; per\; line.\; Each\; query\; starts\; with\; its\; type 11 or 22.\; Each\; query\; of\; the\; first\; type\; continues\; with\; two\; integers ff and tt (1\le f,t\le n1\le f,t\le n) —\; the\; vertex\; the\; pin\; is\; taken\; from,\; and\; the\; vertex\; the\; pin\; is\; put\; to\; and\; the\; polygon\; finishes\; rotating.\; It\; is\; guaranteed\; that\; the\; vertex ff contains\; a\; pin.\; Each\; query\; of\; the\; second\; type\; continues\; with\; a\; single\; integer vv (1\le v\le n1\le v\le n) —\; the\; vertex\; the\; coordinates\; of\; which\; Hag\; should\; tell\; his\; father.}$

It is guaranteed that there is at least one query of the second type.

Output

The output should contain the answer to each query of second type — two numbers in a separate line. Your answer is considered correct, if its absolute or relative error does not exceed ${\mathrm{10-410-4.}}^{}$

Formally, let your answer be $\mathrm{aa,\; and\; the\; jury\text{'}s\; answer\; be bb.\; Your\; answer\; is\; considered\; correct\; if |a-b|max(1,|b|)\le 10-4|a-b|max(1,|b|)\le 10-4}$

Examples

input

Copy

3 4
0 0
2 0
2 2
1 1 2
2 1
2 2
2 3

output

Copy

3.4142135624 -1.4142135624
2.0000000000 0.0000000000
0.5857864376 -1.4142135624

input

Copy

3 2
-1 1
0 0
1 1
1 1 2
2 1

output

Copy

1.0000000000 -1.0000000000

Note

In the first test note the initial and the final state of the wooden polygon.

Red Triangle is the initial state and the green one is the triangle after rotation around $(2,0)(2,0).$

In the second sample note that the polygon rotates $\mathrm{180180 degrees\; counter-clockwise\; or\; clockwise\; direction\; (it\; does\; not\; matter),\; because\; Hag\text{'}s\; father\; makes\; sure\; that\; the\; polygon\; is\; stable\; and\; his\; son\; does\; not\; trick\; him.}$

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const long double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const long double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int n, q;
struct point {
    long double x, y;
    point() {}
    point (long double x, long double y) : x(x), y(y) {}
    void input() {
        //scanf("%lf%lf", &x, &y);
        cin >> x >> y;
    }
    void output() {
        double tx = x, ty = y;
        printf("%.12f %.12f
", tx, ty);
    }
    point operator - (const point &p) const {
        return point(x - p.x, y - p.y);
    }
    point operator + (const point &p) const {
        return point(x + p.x, y + p.y);
    }
    long double operator ^ (const point &p) const {
        return x * p.y - y * p.x;
    }
    point operator / (const long double &k) const {
        return point(x / k, y / k);
    }
    point operator * (const long double &k) const {
        return point(x * k, y * k);
    }
    long double dis(const point &p) const {
        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
    }
} p[100015];
long double d0[100015], d1[100015], th[100015];
long double cal_th(point p1, point p0, point p2) {
    long double th1 = atan2(p1.y - p0.y, p1.x - p0.x);
    long double th2 = atan2(p2.y - p0.y, p2.x - p0.x);
    long double res = th1 - th2;
    if (res < 0) res += 2 * pi;
    return res;
}
multiset<int> rem;
point get_pos(int index) {
    long double th1 = atan2(p[1].y - p[0].y, p[1].x - p[0].x);
    long double th2 = th1 - th[index];
    if (th2 < 0) th2 += 2 * pi;
    return p[0] + point(d0[index] * cos(th2), d0[index] * sin(th2));
}
void rot_and_update(int index2) {
    point p2 = get_pos(index2);
    long double th01 = atan2(p[0].y - p2.y, p[0].x - p2.x);
    long double th02 = 3 * pi / 2;
    long double th0 = th02 - th01;
    if (th0 < 0) th0 += 2 * pi;
    long double th11 = atan2(p[1].y - p2.y, p[1].x - p2.x);
    //p[0].output(), p[1].output(), p2.output();
    long double th12 = th11 + th0;
    if (2 * pi <= th12) th12 -= 2 * pi;
    //printf("index2 = %d
", index2);
    p[0] = point(p2.x, p2.y - d0[index2]);
    p[1] = p2 + point(d1[index2] * cos(th12), d1[index2] * sin(th12));
}
void get_wp() {
    long double s = 0;
    p[0] = point(0, 0);
    for (int i = 2; i < n; ++i) {
        long double ts = (p[i] - p[1]) ^ (p[i] - p[i + 1]);
        p[0] = p[0] + (p[1] + p[i] + p[i + 1]) / 3 * ts;
        s += ts;
    }
    p[0] = p[0] / s;
}
int main() {
    scanf("%d%d", &n, &q);
    p[0] = point(0, 0);
    for (int i = 1; i <= n; ++i) {
        p[i].input();
    }
    get_wp();
    for (int i = 1; i <= n; ++i) {
        th[i] = cal_th(p[1], p[0], p[i]);
        d0[i] = p[0].dis(p[i]);
        d1[i] = p[1].dis(p[i]);
    }
    rem.insert(1), rem.insert(2);
    while (q--) {
        int op;
        scanf("%d", &op);
        if (1 == op) {
            int f, t;
            scanf("%d%d", &f, &t);
            multiset<int>::iterator it = rem.lower_bound(f);
            rem.erase(it);
            int v = *rem.begin();
            rot_and_update(v);
            rem.insert(t);
        } else {
            int v;
            scanf("%d", &v);
            get_pos(v).output();
        }
    }
    return 0;
}
/*
10 10
0 -100000000
1 -100000000
1566 -99999999
2088 -99999997
2610 -99999994
3132 -99999990
3654 -99999985
4176 -99999979
4698 -99999972
5220 -99999964
1 2 5
2 1
1 1 7
2 5
1 5 4
1 4 2
2 8
1 7 9
2 1
1 2 10
*/