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.
a, b = [int(x) for x in input().split()]
lines = []
for i in range(a):
	lines.append(input())
count = 0
face = {'f', 'a', 'c', 'e'}
for i in range(a):
	for j in range(b):
		if(lines[i][j] == 'f'):
			if(i != 0 and j != 0 and {lines[i][j-1], lines[i-1][j], lines[i-1][j-1], lines[i][j]} == face):
				count += 1
			if(i != 0 and j != b-1 and {lines[i][j+1], lines[i-1][j], lines[i-1][j+1], lines[i][j]} == face):
				count += 1
			if(i != a-1 and j != 0 and {lines[i][j-1], lines[i+1][j], lines[i+1][j-1], lines[i][j]} == face):
				count += 1
			if(i != a-1 and j != b-1 and {lines[i][j+1], lines[i+1][j], lines[i+1][j+1], lines[i][j]} == face):
				count += 1
print(count)