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.
poss = [
    'face', 'faec', 'fcae', 'fcea',
    'feac', 'feca', 'afce', 'afec',
    'acfe', 'acef', 'aefc', 'aecf',
    'cfae', 'cfea', 'cafe', 'caef',
    'cefa', 'ceaf', 'efac', 'efca',
    'eafc', 'eacf', 'ecfa', 'ecaf'
]

def check(p, i, j, k):
    return p[i][j:j+2] == poss[k][0:2] and p[i+1][j:j+2] == poss[k][2:4]

n, m = map(int, input().split(' '))
p = [input() for _ in range(n)]
ans = 0

for i in range(n - 1):
    for j in range(m - 1):
        for k in range(24):
            if check(p, i, j, k):
                ans += 1

print(ans)
