Codeforces Round #371 (Div. 2) C. Sonya and Queries —— 二进制压缩

题目链接：http://codeforces.com/contest/714/problem/C

C. Sonya and Queries

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today Sonya learned about long integers and invited all her friends to share the fun. Sonya has an initially empty multiset with integers. Friends give her t queries, each of one of the following type:

1.  +  ai — add non-negative integer ai to the multiset. Note, that she has a multiset, thus there may be many occurrences of the same integer.
2.  -  ai — delete a single occurrence of non-negative integer ai from the multiset. It's guaranteed, that there is at least one ai in the multiset.
3. ? s — count the number of integers in the multiset (with repetitions) that match some pattern s consisting of 0 and 1. In the pattern, 0 stands for the even digits, while 1 stands for the odd. Integer x matches the pattern s, if the parity of the i-th from the right digit in decimal notation matches the i-th from the right digit of the pattern. If the pattern is shorter than this integer, it's supplemented with 0-s from the left. Similarly, if the integer is shorter than the pattern its decimal notation is supplemented with the 0-s from the left.

For example, if the pattern is s = 010, than integers 92, 2212, 50 and 414 match the pattern, while integers 3, 110, 25 and 1030 do not.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 100 000) — the number of operation Sonya has to perform.

Next t lines provide the descriptions of the queries in order they appear in the input file. The i-th row starts with a character ci — the type of the corresponding operation. If ci is equal to '+' or '-' then it's followed by a space and an integer ai (0 ≤ ai < 1018) given without leading zeroes (unless it's 0). If ci equals '?' then it's followed by a space and a sequence of zeroes and onse, giving the pattern of length no more than 18.

It's guaranteed that there will be at least one query of type '?'.

It's guaranteed that any time some integer is removed from the multiset, there will be at least one occurrence of this integer in it.

Output

For each query of the third type print the number of integers matching the given pattern. Each integer is counted as many times, as it appears in the multiset at this moment of time.

Examples

input

12
+ 1
+ 241
? 1
+ 361
- 241
? 0101
+ 101
? 101
- 101
? 101
+ 4000
? 0

output

2
1
2
1
1

input

4
+ 200
+ 200
- 200
? 0

output

1

Note

Consider the integers matching the patterns from the queries of the third type. Queries are numbered in the order they appear in the input.

1. 1 and 241.
2. 361.
3. 101 and 361.
4. 361.
5. 4000.

题解： 

由于查询模式中只有01串（代表奇偶），所以输入的数字只需用01记录每一位奇偶性。故可以用二进制对数字进行压缩归类。由于1<<18=26214, 范围不是很大,所以可以用数组记录每个二进制数出现的个数，之后就可以直接修改，查询了。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#include<set>
#define LL long long
#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
#define INF 0x7fffffff
#define LNF ((1LL<<62)-1)
#define mod 1000000007
#define maxn 300000

using namespace std;

LL p[maxn];
char ch, s[50];

int main()
{
    int n,bin;
    scanf("%d",&n);
    for(int i = 0; i<n; i++)
    {
        getchar();
        scanf("%c%s",&ch,s);
        bin = 0;
        for(int j = 0,len = strlen(s); j<len; j++)
        {
            bin <<= 1;
            if((s[j]-'0')&1)
                bin++;
        }

        if(ch=='+')
            p[bin]++;
        else if(ch=='-')
            p[bin]--;
        else
            printf("%lld
",p[bin]);
    }
}