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
#define rep(i,n) for(int i=0;i<n;i++)
int h,w;
char a[411][411];
int dx[]={1,0,-1,0};
int dy[]={0,1,0,-1};
bool used[411][411];
int ans=0;
int white;
int black;
bool allused[411][411];
void dfs(int x,int y){
	used[x][y]=true;
	allused[x][y]=true;
	if(a[x][y]=='.')white++;
	else black++;
	rep(i,4)if(a[x][y]!=a[x+dx[i]][y+dy[i]]&&!used[x+dx[i]][y+dy[i]]&&(a[x+dx[i]][y+dy[i]]=='#'||a[x+dx[i]][y+dy[i]]=='.'))dfs(x+dx[i],y+dy[i]);
}
signed main(){
	cin>>h>>w;
	for(int i=1;i<=h;i++)for(int j=1;j<=w;j++)cin>>a[i][j];
	for(int i=1;i<=h;i++){
		for(int j=1;j<=w;j++){
			if(a[i][j]=='.')continue;
			if(allused[i][j])continue;
			white=0;
			black=0;
			dfs(i,j);
			ans+=white*black;
		}
	}
	cout<<ans<<endl;
	return 0;
}
