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 int long long
typedef pair<int,int> PII;

bool check[400][400];
char mp[400][400];
signed main(){
	int h,w;
	cin>>h>>w;
	for (int i = 0; i < h; ++i){
		for (int j = 0; j < w; ++j){
			cin>>mp[i][j];
		}
	}

	int ans=0;
	int dx[4]={1,-1,0,0};
	int dy[4]={0,0,1,-1};
	for (int i = 0; i < h; ++i){
		for (int j = 0; j < w; ++j){
			if(mp[i][j]=='.')
				continue;
			else if(check[i][j]==false){
				int black=1,white=0;
				queue<PII> que;
				check[i][j]=true;
				que.push(make_pair(i,j));
				while(que.size()){
					PII p = que.front();
					que.pop();
					for (int k = 0; k < 4; ++k){
						int ny=p.first+dy[k];
						int nx=p.second+dx[k];
						if(mp[ny][nx]!=mp[p.first][p.second]&&check[ny][nx]==false&&nx>=0&&nx<w&&ny>=0&&ny<h){
							que.push(make_pair(ny,nx));
							if(mp[ny][nx]=='.'){
								white++;
							}else if(mp[ny][nx]=='#')
								black++;
							check[ny][nx]=true;
						}
					}
				}

				
				ans+=black*white;
			}

		}

	}
	cout<<ans<<endl;
}

