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<stdio.h>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=505;
string s[maxn];
bool mark[maxn][maxn];
ll cnt1=0,cnt2=0,ans=0;
int n,m,dx[maxn]={0,1,-1,0,0},dy[maxn]={0,0,0,-1,1};
void dfs(int x,int y){
	if(s[x][y]=='#')
		cnt1++;
	else	
		cnt2++;
	mark[x][y]=true;
	for(int i=1;i<=4;i++){
		int tox=x+dx[i],toy=y+dy[i];
		if(tox>=1&&tox<=n&&toy>=0&&toy<m&&!mark[tox][toy]&&s[tox][toy]!=s[x][y])
			dfs(tox,toy);
	}
}
int main(){
	//freopen("","r",stdin);
	//freopen("","w",stdout);
	ios::sync_with_stdio(false);
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		cin>>s[i];
	memset(mark,false,sizeof(mark));
	for(int i=1;i<=n;i++)
		for(int j=0;j<m;j++){
			if(!mark[i][j]){
				cnt1=cnt2=0;	
				dfs(i,j);
				ans+=cnt1*cnt2;
				//cout<<cnt1<<" "<<cnt2<<" "<<i<<" "<<j<<endl;
			}
		}
	cout<<ans<<endl;
	return 0;
}
