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 = [int(x) for x in input().split()]
a = []
count = 0
if m <2 or n<2:
    print(count)
else:
    for i in range(n):
        a.append(input())
    for i in range(n-1):
        for j in range(m-1):
            if a[i][j] != "x":
                if a[i][j+1] != "x" and a[i+1][j] != "x" and a[i+1][j+1] != "x":
                    if a[i][j] != a[i][j+1] and a[i][j] != a[i+1][j+1] and a[i][j] != a[i+1][j] and a[i+1][j] != a[i][j+1] and a[i+1][j] != a[i+1][j+1] and a[i][j+1] != a[i+1][j+1]:
                        count += 1
    print(count)