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
from itertools import*
import math
from collections import*
from heapq import*
from bisect import bisect_left,bisect_right
from copy import deepcopy
inf = 10**18
mod = 10**9+7
from functools import reduce
import sys
sys.setrecursionlimit(10**7)

H,W  = map(int,input().split())
S = [input() for i in range(H)]

dirc = [(0,1),(1,0),(0,-1),(-1,0)]
visited = [[False]*W for _ in range(H)]

def bfs(i,j):
    b,w = 0,0
    visited[i][j] = True
    que = deque([(i,j)])

    while que:
        ci,cj = que.popleft()
        if S[ci][cj] =="#":
            b += 1
        else:
            w += 1

        for di,dj in dirc:
            ni,nj = ci+di,cj+dj
            if not(0<=ni<H and 0<=nj<W) or visited[ni][nj]:
                continue
            if S[ci][cj] != S[ni][nj]:
                visited[ni][nj] = True
                que.append((ni,nj))
    return b*w

res = 0
for i in range(H):
        for j in range(W):
            if visited[i][j]:
                continue
            res += bfs(i,j)
print(res)