Educational Codeforces Round 43 D. Degree Set

D. Degree Set

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence of n positive integers d1, d2, ..., dn (d1 < d2 < ... < dn). Your task is to construct an undirected graph such that:

• there are exactly dn + 1 vertices;
• there are no self-loops;
• there are no multiple edges;
• there are no more than 106 edges;
• its degree set is equal to d.

Vertices should be numbered 1 through (dn + 1).

Degree sequence is an array a with length equal to the number of vertices in a graph such that ai is the number of vertices adjacent to i-th vertex.

Degree set is a sorted in increasing order sequence of all distinct values from the degree sequence.

It is guaranteed that there exists such a graph that all the conditions hold, and it contains no more than 106 edges.

Print the resulting graph.

Input

The first line contains one integer n (1 ≤ n ≤ 300) — the size of the degree set.

The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 1000, d1 < d2 < ... < dn) — the degree set.

Output

In the first line print one integer m (1 ≤ m ≤ 106) — the number of edges in the resulting graph. It is guaranteed that there exists such a graph that all the conditions hold and it contains no more than 106 edges.

Each of the next m lines should contain two integers vi and ui (1 ≤ vi, ui ≤ dn + 1) — the description of the i-th edge.

Examples

input

Copy

3
2 3 4

output

Copy

8
3 1
4 2
4 5
2 5
5 1
3 2
2 1
5 3

input

Copy

3
1 2 3

output

Copy

4
1 2
1 3
1 4
2 3

思路：对于当前度序列（d1, d2, d3, ..., d(k - 1), dk）且总点数为dk + 1，我们取d1个点连接所有的点，取dk - dk - 1个点只连之前这d1个点。此时问题变为d1个度为dk的点，dk - dk -1个度为d1的点，以及待处理度序列
（d2 - d1, d3 - d1, ..., d(k - 1) - d1）,且总点数为dk + 1 - d1 - (dk - d(k - 1)) = d(k - 1) - d1 + 1.刚好是原问题的一个子问题。
（看到这个官方题解的时候很震撼，分析好久才实现出来，但还是想不到）

#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 double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const 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, d[305];
int p[1015];
struct edge {
    int u, v;
    edge () {}
    edge (int u, int v) : u(u), v(v) {}
} e[1000015];
int tot, nex[1000015], head[1015];
void addEdge(int u, int v) {
    e[++tot] = edge(u, v), nex[tot] = head[u], head[u] = tot;
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &d[i]);
    }
    for (int i = 1, index2 = d[n] + 1, index1 = 1; i <= n; ++i) {
        if (!d[i]) continue;
        for (int j = index2; j > index2 - d[i]; --j) {
            for (int k = j - 1; k >= index1; --k) {
                addEdge(j, k);
            }
        }
        index2 -= d[i];
        index1 += d[n] - d[n - 1];
        --n;
        for (int j = i + 1; j <= n; ++j) {
            d[j] -= d[i];
        }
    }
    printf("%d
", tot);
    for (int i = 1; i <= tot; ++i) {
        printf("%d %d
", e[i].u, e[i].v);
    }
    return 0;
}