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 <math.h>
using namespace std;
typedef long long int ll;

int h,w;
char s[420][420];
bool used[420][420];

int dy[4]={0,-1,0,1};
int dx[4]={1,0,-1,0};

ll cnt,cnt1;

void dfs(int x,int y){
	if(used[x][y])return ;
	used[x][y]=true;
	if(s[x][y]=='.')cnt++;
	else cnt1++;
	for(int i=0;i<4;i++){
		int nx=x+dx[i],ny=y+dy[i];
		if(nx<0||ny<0||nx>=h||ny>=w) continue;
		if(s[x][y]==s[nx][ny])continue;
		dfs(nx,ny);
	}
}

int main(){
	cin >> h >> w;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			cin >> s[i][j];
		}
	}
	ll ans=0;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			if(used[i][j])continue;
			cnt=cnt1=0;
			dfs(i,j);
			ans+=cnt*cnt1;
		}
	}
	cout << ans << endl;
}