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<algorithm>
#include<vector>
#include<stack>
#include<math.h>
#include<queue>

using namespace std;
typedef long long int lld;
int arr[1000][1000];
int h,w;
char table[1000][1000];
bool visited[1000][1000];
lld b,W;
int Component;
void DFS(int i, int j){
	if(i<0 || i>=h || j<0 || j>=w)return;
	if(visited[i][j] || arr[i][j]!=Component)return;
	visited[i][j]=true;
	if(table[i][j]=='#')b++;
	else W++;
	DFS(i-1,j);
	DFS(i+1,j);
	DFS(i,j-1);
	DFS(i,j+1);
}
int main(){
	cin>>h>>w;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){visited[i][j]=false;
			cin>>table[i][j];
			if(table[i][j]=='#')arr[i][j]=1;
			else arr[i][j]=0;
			arr[i][j]+=i+j;
			arr[i][j]%=2;
		}
	}
	lld ans=0;
	for(int i=0;i<h;i++){
		for(int j=0;j<w;j++){
			//cout<<arr[i][j];
			if(!visited[i][j]){
				Component=arr[i][j];
				b=0;
				W=0;
				DFS(i,j);
				ans+=b*W;
				//cout<<b<<" "<<W<<endl;
			}
		}//cout<<endl;
	}cout<<ans<<endl;
	return 0;
}