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())
icon = [input() for i in range(n)]
faces = 0
for i in range(n-1):
    for j in range(m-1):
        temp = 0
        temp2 = 0
        temp3 = 0
        temp4 = 0
        if icon[i][j] == "f":
            temp += 1
        elif icon[i][j] == "a":
            temp2 += 1
        elif icon[i][j] == "c":
            temp3 += 1
        elif icon[i][j] == "e":
            temp4 += 1
        if icon[i+1][j] == "f":
            temp += 1
        elif icon[i+1][j] == "a":
            temp2 += 1
        elif icon[i+1][j] == "c":
            temp3 += 1
        elif icon[i+1][j] == "e":
            temp4 += 1
        if icon[i][j+1] == "f":
            temp += 1
        elif icon[i][j+1] == "a":
            temp2 += 1
        elif icon[i][j+1] == "c":
            temp3 += 1
        elif icon[i][j+1] == "e":
            temp4 += 1
        if icon[i+1][j+1] == "f":
            temp += 1
        elif icon[i+1][j+1] == "a":
            temp2 += 1
        elif icon[i+1][j+1] == "c":
            temp3 += 1
        elif icon[i+1][j+1] == "e":
            temp4 += 1
        if temp == 1 and temp2 == 1 and temp3 == 1 and temp4 == 1:
            faces += 1
print(faces)
