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.
def face_found(l):
    f1 = f2 = f3 = f4 = 0
    for ii in l:
            if ii == 'f':
                f1 += 1
            if ii == 'a':
                f2 += 1
            if ii == 'c':
                f3 += 1
            if ii == 'e':
                f4 += 1
    if f1==f2==f3==f4==1:
        return 1
    else:
        return 0
n,m = map(int,input().split(' '))
o = 0
ll = []
for i in range(n):
    l = list(input())
    ll.append(l)
for i in range(n-1):
    for ii in range(m-1):
        p = [ll[i][ii],ll[i+1][ii],ll[i][ii+1],ll[i+1][ii+1]]
        if face_found(p):
            o += 1
print(o)
