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.
# print ("Input n and m")
thepair = input().split()
n = int(thepair[0])
m = int(thepair[1])
if (n==1 or m==1):
    print(0)
else:    #  Each dimension at least two
    answer = 0
    a = [[0 for i in range(m)] for j in range(n)]  # Initialize to all zero
    for r in range(n):
        # print ("Input the next row")
        a[r] = input()

    for r in range(0,n-1):
        for c in range(0,m-1):
            s = []
            s.append(a[r][c])
            s.append(a[r+1][c])
            s.append(a[r][c+1])
            s.append(a[r+1][c+1])
            s = sorted(s)
            if (s[0] == "a" and s[1] == "c" and s[2] == "e" and s[3] == "f"):
                answer = answer + 1
    print (answer)
                
    
            
    
