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.

題意： + a 代表在一個集合裡+一個數， - a代表集合裡減掉這個a， ? 010001代表集合裡每個數字a從右往左跟那個01組成的串匹配，是1代表偶數，0代表奇數，如果其中一個短了，用0補齊，輸出現在集合多少個匹配的； 思路：我最開始寫的是模擬。。。真的想他那樣匹配。。。按理說可以a，可能程式碼能力太差，還是沒a，後來知道了，每輸入一個數字，就先按照奇偶性把他轉換成相應的匹配01串，然後當做二進位制算出來他的和，為什麼當做二進位制算出他的和呢，因為每個二進位制數的01串都是唯一確定的啊，所以搜尋的時候可以直接用啊，然後用一個數組記錄這個二進位制值得數量，如果是  這樣也不用管是不是哪個數比哪個數短補零問題了。。

#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;
}