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.
n, m = map(int, input().split())
L = ['a', 'c', 'e', 'f']
L2 = []
ch = []
nbr = 0

if n == 1 or m == 1:
    print(nbr)
else:
    for i in range(n):
        y = input()
        ch.append(y)
    
    if n >= 2 and m >= 2:
        for i in range(n-1):
            for j in range(m-1):
                if ch[i][j] in L and ch[i][j+1] in L and ch[i+1][j] in L and ch[i+1][j+1] in L:
                    L2 = [ch[i][j], ch[i][j+1], ch[i+1][j], ch[i+1][j+1]]
                    if sorted(L2) == L:
                        nbr += 1
            
    print(nbr)

  