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<bits/stdc++.h>
#define mx 405
using namespace std;

int h, w, dx[] = {0, -1, 0, 1}, dy[] = {1, 0, -1, 0};
bool used[mx][mx], sp[mx][mx];
string s[mx];
long long ans;

int main(){
	scanf("%d%d", &h, &w);
	for(int i = 0; i < h; i++){
		cin >> s[i];
		for(int j = 0; j < w; j++) sp[i][j] = ((s[i][j] == '#') + i + j)&1;
	}

	for(int i = 0; i < h; i++){
		for(int j = 0; j < w; j++){
			if(used[i][j]) continue;
			used[i][j] = true;
			long long cnt = 1;
			long long cn = (s[i][j] == '#');
			stack< pair<int, int> > st;
			st.push(make_pair(i, j));
			while(!st.empty()){
				auto stt = st.top();
				st.pop();
				for(int k = 0; k < 4; k++){
					int di = stt.first + dx[k], dj = stt.second + dy[k];
					if(0 <= di && di < h && 0 <= dj && dj < w && !used[di][dj] && (sp[i][j] == sp[di][dj])){
						used[di][dj] = true;
						st.push(make_pair(di, dj));
						cnt++;
						if(s[di][dj] == '#') cn++;
					}
				}
			}
			ans += (cnt-cn)*cn;
		}
	}
	printf("%lld\n", ans);
}