CF

1.题目描述：

A. Face Detection

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The developers of Looksery have to write an efficient algorithm that detects faces on a picture. Unfortunately, they are currently busy preparing a contest for you, so you will have to do it for them.

In this problem an image is a rectangular table that consists of lowercase Latin letters. A face on the image is a 2 × 2 square, such that from the four letters of this square you can make word "face".

You need to write a program that determines the number of faces on the image. The squares that correspond to the faces can overlap.

Input

The first line contains two space-separated integers, n and m (1 ≤ n, m ≤ 50) — the height and the width of the image, respectively.

Next n lines define the image. Each line contains m lowercase Latin letters.

Output

In the single line print the number of faces on the image.

Examples

input

4 4xxxxxfaxxcexxxxx

output

1

input

4 2xxcfaexx

output

1

input

2 3faccef

output

2

input

1 4face

output

0

Note

In the first sample the image contains a single face, located in a square with the upper left corner at the second line and the second column:

In the second sample the image also contains exactly one face, its upper left corner is at the second row and the first column.

In the third sample two faces are shown:

In the fourth sample the image has no faces on it.

2.题意概述：

有一个face的定义为某2x2矩阵内有face字母，不论顺序

3.解题思路：

矩阵长宽小于50，直接暴力模拟就行

4.AC代码：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 100100#define N 55#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;char mp[N][N];int main(){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();#endifint n, m;while (~scanf("%d%d", &n, &m)){for (int i = 0; i < n; i++)scanf("%s", mp[i]);int cnt = 0;for (int i = 0; i < n - 1; i++)for (int j = 0; j < m - 1; j++){int f1 = 0, f2 = 0, f3 = 0, f4 = 0;for (int k = 0; k < 2; k++)for (int l = 0; l < 2; l++){if (mp[i + k][j + l] == 'f')f1++;else if (mp[i + k][j + l] == 'a')f2++;else if (mp[i + k][j + l] == 'c')f3++;else if (mp[i + k][j + l] == 'e')f4++;}if (f1 && f2 && f3 && f4)cnt++;}printf("%d\n", cnt);}#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);#endifreturn 0;}