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>
 
using namespace std;
 
#define INF 1000000007
#define LINF 1000000000000000007
 
typedef long long ll;
typedef pair<ll,ll> P;


ll h, w;
string s[404];
bool used[400][400] = {0};

int main(){
	cin >> h >> w;
	for(int i = 0; i < h; i++){
		cin >> s[i];
	}
	
	ll ans = 0;
	for(int i = 0; i < h; i++)for(int j = 0; j < w; j++){
		if(!used[i][j]){
			ll x = 0, y = 0;
			queue<P> que;
			que.push(P(i,j));
			used[i][j] = 1;
			while(!que.empty()){
				P q = que.front();
				que.pop();
				if(s[q.first][q.second] == '#') x++;
				else y++;
				int dx[4] = {0,-1,0,1}, dy[4] = {-1,0,1,0};
				for(int j = 0; j < 4; j++){
					int nx = dx[j]+q.first, ny = dy[j]+q.second;
					if(nx < 0 || nx >= h || ny < 0 || ny >= w) continue;
					if(used[nx][ny]) continue;
					if(s[q.first][q.second] != s[nx][ny]){
						que.push(P(nx,ny));
						used[nx][ny] = 1;
					}
				}
			}
			ans += x * y;
		}
	}
	cout << ans << endl;
	
	return 0;
}