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())
vetor = []
soma = 0
for i in range(n):
    lista = list(input())
    vetor.append(lista)
for i in range(n-1):
    for j in range(m-1):
        if(vetor[i][j]!=vetor[i+1][j]!=vetor[i][j+1]!=vetor[i+1][j+1])and (vetor[i][j] == 'f' or vetor[i][j] == 'a' or vetor[i][j] == 'c' or vetor[i][j] == 'e')and (vetor[i+1][j] == 'f' or vetor[i+1][j] == 'a' or vetor[i+1][j] == 'c' or vetor[i+1][j] == 'e')and (vetor[i][j+1] == 'f' or vetor[i][j+1] == 'a' or vetor[i][j+1] == 'c' or vetor[i][j+1] == 'e')and (vetor[i+1][j+1] == 'f' or vetor[i+1][j+1] == 'a' or vetor[i+1][j+1] == 'c' or vetor[i+1][j+1] == 'e'):
                soma += 1;
print(soma)
 	 	  	  			   			    		   		 	