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

for i in range(m):
    image.append(" ".join(input()).split())

col, row = 0, 0

def is_face(square):
    face = ['f', 'a', 'c', 'e']
    char_count = 0
    for i in range(2):
        for j in range(2):
            if square[i][j] != "x":
                char_count += 1
                face.remove(square[i][j])
    if char_count > 1:
        if not face:
            return 2
        else:
            return 1
    else:
        return 0

face_count = 0

while row != m-1:
    square = [image[row][col:col+2], image[row+1][col:col+2]]
    offset = 1
    face = is_face(square)
    if face == 2:
        face_count += 1
    elif not face:
        offset = 2

    if col + offset > n - 2:
        col = 0
        row += 1
    else:
        col += offset

print(face_count)