题干：

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

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

Output

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 .

题目大意：

维护一个初始时只有元素0的多重集，有三种操作

1  x: 添加一个数x

2  x: 删除一个数x

3  x: 询问一个数x与这个集合中的某一个元素的异或值最大是多少。

总共q次操作，每次操作是xi  (1 ≤ xi ≤ 1e9,1 ≤ q ≤ 200 000)

解题报告：

直接01字典树，注意刚开始需要先插入0.再就是数组需要开大一些左右，2e5是不够的。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1e7 + 5;
int tr[MAX][2],tot,num[MAX];
void ins(int x) {int rt = 0;for(int bit = 31; bit>=0; bit--) {int tar = (x&(1<<bit)) > 0 ? 1 : 0;if(tr[rt][tar] == 0) tr[rt][tar] = ++tot;rt=tr[rt][tar]; num[rt]++;}
}
void del(int x) {int rt = 0;for(int bit = 31; bit>=0; bit--) {int tar = (x&(1<<bit)) > 0 ? 1 : 0;rt=tr[rt][tar]; num[rt]--;}
}
int cal(int x) {int res = 0,rt = 0;for(int bit = 31; bit>=0; bit--) {int tar = (x&(1<<bit)) > 0 ? 1 : 0;if(num[tr[rt][!tar]] > 0) res |= (1<<bit),rt=tr[rt][!tar];else rt = tr[rt][tar];if(rt == 0) break;} return res;
}
int main()
{int q,x; char op[5];cin>>q;ins(0);while(q--) {scanf("%s%d",op,&x);if(op[0] == '+') ins(x);if(op[0] == '-') del(x);if(op[0] == '?') printf("%d\n",cal(x));}return 0 ;
}