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.
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.
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

def test(f, a, c, e):
    face = [f, a, c, e]
    if 'f' in face and 'a' in face and 'c' in face and 'e' in face:
        return 1
    return 0

import time

(n, m) = (int(i) for i in input().split())
fmap   = []

for i in range(n):
    fmap.append(input())

start = time.time()
acc   = 0
for i in range(n):
    for j in range(m):
        if fmap[i][j] == 'f':
            if i< n-1 and j <m-1:
                acc += test(fmap[i][j], fmap[i][j+1], fmap[i+1][j], fmap[i+1][j+1])
            if i < n-1 and j > 0:
                acc += test(fmap[i][j], fmap[i][j-1], fmap[i+1][j], fmap[i+1][j-1])
            if i > 0 and j <m-1:
                acc += test(fmap[i][j], fmap[i][j+1], fmap[i-1][j], fmap[i-1][j+1])
            if i > 0 and j > 0:
                acc += test(fmap[i][j], fmap[i][j-1], fmap[i-1][j], fmap[i-1][j-1])

print(acc)
finish = time.time()
#print(finish - start)

