codeforces 894 A题 QAQ

A. QAQ

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).

Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.

Input

The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.

Output

Print a single integer — the number of subsequences "QAQ" in the string.

Examples

Input

QAQAQYSYIOIWIN

Output

4

Input

QAQQQZZYNOIWIN

Output

3

Note

In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

 题意：给一个长度 不超过 100 的字符串 ， 统计 QAQ 的出现次数 ， （QAQ的出现次数 ：QAQ 不一定要是连续的 ， 但是位置上 一定是 A的两边各有一个Q）

思路：前缀和 计算 第 i 个位置 和之前一共出现了多少个 Q ， 则遍历枚举 A 累加 A之前 和 A之后 的零的个数的乘积。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std ;
 
#define LL long long 
#define maxn 200 
char str[maxn] ; 
int sum[maxn] ; 

int main(){
    
    while(~scanf(" %s" , str )){
        memset(sum , 0 , sizeof(sum)) ; 
        int num = 0 ; 
        int len = strlen(str) ; 
        for(int i=0 ; i<len ; i++){
            if(str[i] =='Q'){
                num ++ ; 
            }
            
            sum[i] = num ; 
        }
        LL result = 0 ; 
        for(int i=0 ; i<len ; i++){
            if(str[i] =='A')
            result += sum[i] * (sum[len - 1]-sum[i]) ; 
        }
        
        printf("%lld\n" , result) ; 
    }
    
    return 0 ;
}