D. Vasiliy's Multiset 异或字典树

D. Vasiliy's Multiset

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Author has gone out of the stories about Vasiliy, so here is just a formal task description.

You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:

1. "+ x" — add integer x to multiset A.
2. "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
3. "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.

Multiset is a set, where equal elements are allowed.

Input

The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.

Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.

Note, that the integer 0 will always be present in the set A.

Output

For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.

Example

input

Copy

10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11

output

Copy

11
10
14
13

Note

After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.

The answer for the sixth query is integer  — maximum among integers , , , and .

就是有3种操作

+ x向集合里添加 x

- x 删除x元素,(保证存在

? x 查询 x |  集合中元素的最大值

新姿势 get，就是利用字典树，从高位到低位进行贪心。想到从高位求，但是不知道怎么用字典树..orz.新姿势

#include <algorithm>
#include <stack>
#include <istream>
#include <stdio.h>
#include <map>
#include <math.h>
#include <vector>
#include <iostream>
#include <queue>
#include <string.h>
#include <set>
#include <cstdio>
#define FR(i,n) for(int i=0;i<n;i++)
#define MAX 2005
#define mkp pair <int,int>
using namespace std;
//#pragma comment(linker, "/STACK:10240000000,10240000000")
const int maxn = 4e6+4;
typedef long long ll;
const int  inf = 0x3fffff;
void read(int  &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
    x *= 1 - 2 * flag;
}

int tree[maxn][3];
int sum[maxn];

int cnt = 2;
void Insert(int val,ll t)
{
    int root = 1;
    for(int i=31;i>=0;i--)
    {
        int id=(t>>i)&1;
        if(!tree[root][id])
        {
            tree[root][id]=cnt++;
        }
        root = tree[root][id];
        sum[root]+=val;
    }
}

ll ans(ll tmp)
{
    tmp=~tmp;
    int root = 1;
    ll res = 0;
    for(int i=31;i>=0;i--)
    {
        int id = (tmp>>i)&1;
        res*=2;
        if(tree[root][id]&&sum[tree[root][id]])
        {
            res++;
            root = tree[root][id];
        }
        else
        {
            root=tree[root][1-id];
        }
    }
    return res;
}

int main() {
    int n;
    read(n);
    memset(sum,0,sizeof(sum));
    memset(tree,0,sizeof(tree));
    Insert(1,0);
    /*for(int i=0;i<32;i++){
        printf("%d %d
",tree[i][0],tree[i][1]);
    }*/
    for(int i=0;i<n;i++){
        char c;
        ll t;
        cin>>c>>t;
        if(c=='+')Insert(1,t);
        else if(c=='-')Insert(-1,t);
        else {
            printf("%lld
",ans(t));
        }
    }
    return 0;
}