Codeforces 894.A QAQ

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

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!).

illustration by 猫屋 https://twitter.com/nekoyaliu

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

input

QAQQQZZYNOIWIN

output

Note

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

题目大意：找字符串QAQ出现的次数.

分析：题目数据范围不大，三种暴力循环就可以过，事实上有O(n)的写法，设f[i][1/2/3]分别为前i位匹配了1/2/3位的方案数，对于每一位判断一下是否等于Q或A,非常简单地转移一下就好了.

#include <cstdio>

#include <cstring>

#include <iostream>

#include <stack>

#include <algorithm>

using namespace std;

char s[];

int f[][], len;

int main()

{

    scanf("%s", s + );

    len = strlen(s + );

    for (int i = ; i <= len; i++)

    {

        f[i][] = f[i - ][];

        f[i][] = f[i - ][];

        f[i][] = f[i - ][];

        if (s[i] == 'Q')

        {

            f[i][]++;

            f[i][] += f[i - ][];

        }

        if (s[i] == 'A')

            f[i][] += f[i - ][];

    }

    printf("%d\n", f[len][]);

return ;

}