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.
"""
Codeforces Looksery Cup 2015 Problem A

Author  : chaotic_iak
Language: Python 3.4.2
"""

################################################### SOLUTION

def main():
    n,m = read()
    s = []
    for i in range(n): s.append(read(0))

    ans = 0
    for i in range(n-1):
        for j in range(m-1):
            if {s[i][j], s[i][j+1], s[i+1][j], s[i+1][j+1]} == set("face"):
                ans += 1
    return ans



#################################################### HELPERS



def read(mode=2):
    # 0: String
    # 1: List of strings
    # 2: List of integers
    inputs = input().strip()
    if mode == 0: return inputs
    if mode == 1: return inputs.split()
    if mode == 2: return list(map(int, inputs.split()))

def write(s="\n"):
    if s is None: s = ""
    if isinstance(s, list): s = " ".join(map(str, s))
    s = str(s)
    print(s, end="")

write(main())