Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <utility>
#define maxn 405
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int h,w;
char s[maxn][maxn];
bool vis[maxn][maxn];
const int dx[] = {-1,0,1,0};
const int dy[] = {0,1,0,-1};
int numb = 0,numw = 0;
void dfs(int x,int y){
	vis[x][y] = true;
	if(s[x][y] == '#') numb++;
	else numw++;
	for(int d = 0;d < 4;d++){
		int cx = x + dx[d], cy = y + dy[d];
		if(cx >= 1 && cx <= h && cy >= 1 && cy <= w && !vis[cx][cy] && s[cx][cy] != s[x][y]){
			dfs(cx,cy);
		}
	}
}
int main(){
	scanf("%d%d",&h,&w);
	for(int i=1;i<=h;i++){
		scanf("%s",s[i] + 1);
	}
	ll ans = 0;
	for(int i=1;i<=h;i++){
		for(int j=1;j<=w;j++){
			if(!vis[i][j]){
				numb = numw = 0;
				dfs(i,j);
				ans += (ll)numb * numw;
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}