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 = str(input()).split(' ')
n = int(n)
m = int(m)

matrix = []

for i in range(n):
    matrix.append(list(input()))

counter = 0
for i in range(n-1):
    for j in range(m-1):
        if matrix[i][j] == "f" or matrix[i][j] == "a" or matrix[i][j] == "c" or matrix[i][j] == "e":
            word = matrix[i][j]
            if matrix[i][j+1] == "f" or matrix[i][j+1] == "a" or matrix[i][j+1] == "c" or matrix[i][j+1] == "e":
                word += matrix[i][j+1]
                if matrix[i+1][j] == "f" or matrix[i+1][j] == "a" or matrix[i+1][j] == "c" or matrix[i+1][j] == "e":
                    word += matrix[i+1][j]
                    if matrix[i+1][j+1] == "f" or matrix[i+1][j+1] == "a" or matrix[i+1][j+1] == "c" or matrix[i+1][j+1] == "e":
                        word += matrix[i+1][j+1]
                        if ("f" in word) and ("a" in word) and ("c" in word) and ("e" in word):
                            counter += 1
print(counter)