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())
a = [ [0] * (m) for i in range(n) ]
for i in range(n):
    tmp = input()
    for j in range(len(tmp)):
        a[i][j] = tmp[j]
set_ = set()
s = 0
e = 0
for i in range(len(a)):
    for j in range(len(a[i])):
        set_.add(a[i][j])
        s = s + 1
        if (j + 1 < m):
            if ((a[i][j+1] == 'e') or (a[i][j+1] == 'a') or (a[i][j+1] == 'c') or (a[i][j+1] == 'f')):
                if (a[i][j+1] not in set_):
                    s = s + 1
                    set_.add(a[i][j+1])
        if ((i+1 < n) and (j + 1 < m)):
            if ((a[i+1][j+1] == 'e') or (a[i+1][j+1] == 'c' or (a[i+1][j+1] == 'f') or (a[i+1][j+1] == 'a'))):
                if (a[i+1][j+1] not in set_):
                    s = s + 1
                    set_.add(a[i+1][j+1])
        if (i + 1 < n):
            if ((a[i+1][j] == 'a') or (a[i+1][j] == 'e') or (a[i+1][j] == 'c') or (a[i+1][j] == 'f')):
                if (a[i+1][j] not in set_):
                    s = s + 1
                    set_.add(a[i+1][j])
        if (s == 4):
            e = e + 1
        s = 0
        set_ = set()
print(e)