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=[]
for i in range(n):
    a.append(list(input()))
cnt1=0
cnt2=0
cnt3=0
cnt4=0
cnt=0
for i in range(n):
    for j in range(m):
        if i+1<n and j+1<m:
            if a[i][j]=='f' or a[i+1][j]=='f' or a[i][j+1]=='f' or a[i+1][j+1]=='f':
                cnt1=cnt1+1
            if a[i][j]=='a' or a[i+1][j]=='a' or a[i][j+1]=='a' or a[i+1][j+1]=='a':
                cnt2=cnt2+1
            if a[i][j]=='c' or a[i+1][j]=='c' or a[i][j+1]=='c' or a[i+1][j+1]=='c':
                cnt3=cnt3+1
            if a[i][j]=='e' or a[i+1][j]=='e' or a[i][j+1]=='e' or a[i+1][j+1]=='e':
                cnt4=cnt4+1
            if(cnt1>0 and cnt2>0 and cnt3>0 and cnt4>0):
                cnt=cnt+1
            cnt1=0
            cnt2=0
            cnt3=0
            cnt4=0
print(cnt)
            

  


    