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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define LL long long
int H, W;
char S[410][410];
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
bool used[410][410];
void dfs(int x, int y, int &b, int &w){
	used[x][y] = true;
	if(S[x][y] == '#') b++;
	else w++;
	for(int i = 0; i < 4; i++){
		int rx = x + dx[i];
		int ry = y + dy[i];
		if(1 <= rx && rx <= H && 1 <= ry && ry <= W){
			if(S[x][y] != S[rx][ry] && !used[rx][ry]){
				dfs(rx, ry, b, w);
			}
		}
	}
}


int main(){
	cin >> H >> W;
	for(int i = 1; i <= H; i++){
		for(int j = 1; j <= W; j++){
			cin >> S[i][j];
		}
	}
	LL ans = 0;
	for(int i = 1; i <= H; i++){
		for(int j = 1; j <= W; j++){
			int b, w;
			b = w = 0;
			if(!used[i][j]){
				dfs(i, j, b, w);
			}
			ans += (LL)b * (LL)w;
		}
	}
	cout << ans << endl;
	return 0;
}