題目大意

給出一些數字，詢問一些滿足對應位為偶數(0表示)或奇數(1表示)的數字的個數。
比如010格式的數字可以有818,52,98,2212等。

題目翻譯

今天Sonya學到了長整數並且邀請了她的朋友分享喜悅之情。Sonya一開始有一個空的存整數的多重集合。朋友們給她t(1≤t≤100000)個詢問，每一個詢問有以下幾種型別：
1. + ai – 新增一個非負整數ai(0≤ai≤1018)到多重集合中。注意，她有一個多重集合，故集合中1個整數可以出現多次。
2. - ai – 刪掉一個在多重集合中已經有的非負整數ai。保證該整數存在。
3. ? s(length(s)≤18) – 求多重集合中特定整數的個數（重複的也要算），特定整數指的是由0和1組成的模板s，在s中，0表示偶數，1表示奇數。一個整數x匹配模板s僅當x從右數起的第i位數字符合模板s從右數起的第i位。如果模板s比x短，可以在s前加一些0。類似的，如果x比s短，也可以在x前加一些0。保證至少有一個‘？’型詢問。

比如：如果模板s=010，那麼92,2122,50,414都是匹配s的，而3,110,25,1030不匹配。

法1

首先各位數具體為多少並沒有什麼關係，所以全都模2，將長度擴充套件到18方便位數不相等的查詢。
然後扔到字典樹裡查詢就好啦。

法2

9.25
後來又覺得自己好蠢。。
開個陣列不就好了。。字典樹什麼的根本不需要啊。

沒方法法

#include <cstdio>
typedef long long ll;
const int N = 262144;
int c[N];
int main() {
    int t, i, p; ll d; char op;
    scanf
 
("%d", &t);
    while (t--) {
        for (op = getchar(); op == '\n'; op = getchar());
        scanf("%I64d", &d);
        for (i = p = 0; i < 18; ++i)
            p |= ((d % 10) & 1) << i, d /= 10;
        switch(op) {
        case '+': ++c[p]; break;
        case '-': --c[p]; break
 
;
        case '?': printf("%d\n", c[p]); break;
        }
    }
    return 0;
}

字典樹法

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 500005;
int cnt = 0;
int c[N][2], val[N];
void modify(char *s, int d) {
    int i, p = 0, x, l = strlen(s);
    reverse(s, s + l);
    for (; l < 19; ++l) s[l] = '0';
    for (i = 0; i < l; ++i) {
        x = (s[i] - '0') & 1;
        if (!c[p][x]) c[p][x] = ++cnt;
        p = c[p][x]; val[p] += d;
    }
}
int query(char *s) {
    int i, b = 0, j, p = 0, x, l = strlen(s);
    reverse(s, s + l);
    for (; l < 19; ++l) s[l] = '0';
    for (i = 0; i < l; ++i) {
        x = (s[i] - '0') & 1;
        if (!c[p][x]) return 0;
        p = c[p][x];
    }
    return val[p];
}
char valid_char() {
    char ch = getchar();
    while (ch == ' ' || ch == '\n') ch = getchar();
    return ch;
}
char d[N];
int main() {
    int t; char op;
    scanf("%d", &t);
    while (t--) {
        op = valid_char(); scanf("%s", d);
        switch(op) {
        case '+': modify(d, 1); break;
        case '-': modify(d, -1); break;
        case '?': printf("%d\n", query(d)); break;
        }
    }
    return 0;
}

A. Sonya and Queries

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:

 +  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.
 -  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.
? 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 and 241.
361.
101 and 361.
361.
4000.