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.
inp = input().split(" ")
ny = int(inp[0])
nx = int(inp[1])

grid=[]

for y in range(ny):
    grid.append([x for x in input()])

count = 0

for oy in range(ny-1):
    for ox in range(nx-1):
        has_f = False
        has_a = False
        has_c = False
        has_e = False
        for y in range(0, 2):
            for x in range(0, 2):
                item = grid[oy+y][ox+x]
                if item == "f":
                    has_f = True
                elif item == "a":
                    has_a = True
                elif item == "c":
                    has_c = True
                elif item == "e":
                    has_e = True
        if has_f and has_a and has_c and has_e:
            count += 1

print(count)
    
  	 	 	  	   	 		 	 			  		  		