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())
maps = [
    list("face"),
    list("faec"),
    list("fcae"),
    list("fcea"),
    list("feac"),
    list("feca"),
    list("afce"),
    list("afec"),
    list("acfe"),
    list("acef"),
    list("aefc"),
    list("aecf"),
    list("cfae"),
    list("cfea"),
    list("cafe"),
    list("caef"),
    list("cefa"),
    list("ceaf"),
    list("efac"),
    list("efca"),
    list("eafc"),
    list("eacf"),
    list("ecfa"),
    list("ecaf")]
ma = []
for i in range(n):
    *s, = input()
    ma.append(s)
a = 0
for i in range(n - 1):
    for j in range(m - 1):
        if any(x[0] == ma[i][j] and x[1] == ma[i + 1][j] and x[2] == ma[i][j + 1] and x[3] == ma[i + 1][j + 1] for x in maps):
            a += 1
print(a)
