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
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iostream>

using namespace std;

int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};
string s[400];
int H,W;
long color[2];
bool list[400][400];

void dfs(int x, int y){
	char now = s[x][y];

	list[x][y] = false;

	for(int i=0;i<4;i++){
		if(x + dx[i] >= 0 && x + dx[i] < H && y + dy[i] >= 0  && y + dy[i] < W && list[x+dx[i]][y+dy[i]] && s[x+dx[i]][y+dy[i]] != now){
			if(s[x+dx[i]][y+dy[i]] == '#') color[0] += 1;
			else color[1] += 1;
			dfs(x+dx[i],y+dy[i]);
		}
	}
}

int main(void){
	cin >> H >> W;
	for(int i=0; i<H; i++){
		cin >> s[i];
	}
	for(int i=0; i<H; i++){
		for(int j=0; j<W; j++){
			list[i][j] = true;
		}
	}
	long ans =0;

	for(int i=0; i<H; i++){
		for(int j=0; j<W; j++){
			if(s[i][j] == '#' && list[i][j]){
				color[0] = 1;
				color[1] = 0;
				dfs(i,j);
				ans += color[0] * color[1];
			}
		}
	}

	cout << ans << endl;
	return 0;
}

