Codeforces Round #371 (Div. 2）C. Sonya and Queries

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, 0stands 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 with0-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.

题意： + a 代表在一个集合里+一个数， - a代表集合里减掉这个a， ? 010001代表集合里每个数字a从右往左跟那个01组成的串匹配，是1代表偶数，0代表奇数，如果其中一个短了，用0补齐，输出现在集合多少个匹配的；

思路：我最开始写的是模拟。。。真的想他那样匹配。。。按理说可以a，可能代码能力太差，还是没a，后来知道了，每输入一个数字，就先按照奇偶性把他转换成相应的匹配01串，然后当做二进制算出来他的和，为什么当做二进制算出他的和呢，因为每个二进制数的01串都是唯一确定的啊，所以搜索的时候可以直接用啊，然后用一个数组记录这个二进制值得数量，如果是  这样也不用管是不是哪个数比哪个数短补零问题了。。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1000010;
int ans[maxn];
int main()
{
    int n;
    long long a;
    char ch;
    scanf("%d",&n);
    memset(ans,0,sizeof(ans));
    while(n--)
    {
        scanf(" %c %I64d",&ch, &a);
        int sum = 0, k = 1;
        while(a)
        {
            sum += (a%2)*k;
            a /= 10;
            k *= 2;
        }
        if(ch == '+') ans[sum]++;
        if(ch == '-') ans[sum]--;
        if(ch == '?') cout << ans[sum] << endl;
    }
    return 0;
}