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())
k=["0"*(m+2)]
for i in range(n):
     k.append("0"+input()+"0")
k.append("0"*(m+2))
count=0
for i in range(1,n):
     for j in range(1,m):
          k1=(k[i][j],k[i][j+1],k[i+1][j],k[i+1][j+1])
          if "f" in k1 and "a" in k1 and "c" in k1 and "e" in k1:
               count+=1
print(count)
     

