There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`.

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

* There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

* 1 \leq H, W \leq 400
* |S_i| = W (1 \leq i \leq H)
* For each i (1 \leq i \leq H), the string S_i consists of characters `#` and `.`.

Input

Input is given from Standard Input in the following format:


H W
S_1
S_2
:
S_H


Output

Print the answer.

Examples

Input

3 3
.#.
..#
#..


Output

10


Input

3 3
.#.
..#
..


Output

10


Input

2 4
....
....


Output

0


Input

4 3


...


Output

6
import sys
sys.setrecursionlimit(10**7)
h,w=map(int,input().split())
s=[list(input()) for _ in range(h)]
visited=[[False]*w for _ in range(h)]

def dfs(y,x):
    visited[y][x]=True
    blk,wht=0,0
    for dy,dx in ((-1,0),(0,-1),(1,0),(0,1)):
      ny,nx=y+dy,x+dx
      if (not (0<=ny<h)) or (not (0<=nx<w)) or visited[ny][nx]: 
        continue
      if s[y][x]!=s[ny][nx]:
        nb,nw=dfs(ny,nx)
        blk+=nb
        wht+=nw
    if s[y][x]=="#":
      blk+= 1
    else:
      wht+= 1
    return blk,wht

ans=0
for i in range(h):
  for j in range(w):
    #到達できる限り進み、残ったやつを順にdfs
    if not visited[i][j]:
      blk,wht=dfs(i,j)
      ans+=blk*wht
print(ans)