Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset

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

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 .

真是的，邀请赛的时候见过类似的题目；

字典树、查询的时候贪心的查询、如果当前是0  那么我就尽量去找同层的1  如果当前是0 那么就尽量去找同层的1 .

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/12 10:36:45
File Name     :cf367d.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

struct node{
    int next[2];
    int v;
    void init(){
        v=0;
        memset(next,-1,sizeof next);
    }
};
node L[210000*32];
int tot=0;
void add(char s[],int len){
    int now=0;
    for(int i=0;i<len;i++){
        int t=s[i]-'0';
        int next=L[now].next[t];
        if(next==-1){
            next=++tot;
            L[next].init();
            L[now].next[t]=next;
        }
        now=next;
        L[now].v++;
    }
    //L[now].v++;
}

void dele(char s[],int len){
    int now=0;
    for(int i=0;i<len;i++){
        int t=s[i]-'0';
        int next=L[now].next[t];
        now=next;
        L[now].v--;
    }
//    L[now].v--;
    for(int i=0;i<2;i++)L[now].next[i]=-1;
}
int query(char s[],int len){
    int now=0;
    int ans=0;
    for(int i=0;i<len;i++){
        int t=s[i]-'0';
        int next=L[now].next[1-t];
        if(next==-1||L[next].v<=0){
            //这个点不存在或者已经删除了
            next=L[now].next[t];
        }
        else{
            ans+=pow(2.0,32-i-1);; 
        }
        now=next;
    }
    return ans;
}
void cha(int x,char *s){
    for(int i=0;i<=31;i++){
        s[i]=x%2+'0';
        x/=2;
    }
    s[32]='';
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    char s[32],c;
    int n,d;
    cin>>n;
    L[0].init();
    cha(0,s);
    add(s,32);
    for(int i=1;i<=n;i++){    
        cin>>c>>d;
        cha(d,s);
        reverse(s,s+32);
        if(c=='+'){
            add(s,32);
        }
        else if(c=='-'){
            dele(s,32);
        }
        else{
            printf("%d
",query(s,32));
        }
    }
    return 0;
}