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 <vector>
#include <string>
#include <utility>

using namespace std;
string s[500];
vector<int> G[200000];
bool used[200000];
long long a[200000],b[200000] = {},wh[200000] = {};
void dfs(int x){
	used[x] = true;
	for(int v:G[x]){
		if(!used[v]){
			a[v] = a[x];
			dfs(v);
		}
	}
}

int main(){
	int i,j,h,w,n;
	cin >> h >> w;
	n = h*w;
	for(i=0;i<h;i++){
		cin >> s[i];
	}
	for(i=0;i<h;i++){
		for(j=0;j<w;j++){
			if(j!=w-1 && s[i][j]!=s[i][j+1]){
				G[i*w+j].push_back(i*w+j+1);
				G[i*w+j+1].push_back(i*w+j);
			}
			if(i!=h-1 && s[i][j]!=s[i+1][j]){
				G[i*w+j].push_back((i+1)*w+j);
				G[(i+1)*w+j].push_back(i*w+j);
			}
		}
	}
	for(i=0;i<n;i++) used[i] = false;
	for(i=0;i<n;i++){
		if(!used[i]){
			a[i] = i;
			dfs(i);
		}
	}
	for(i=0;i<n;i++){
		if(s[i/w][i%w]=='#') b[a[i]]++;
		else wh[a[i]]++;
	}
	long long ans = 0;
	for(i=0;i<n;i++){
		ans += b[i]*wh[i];
	}
	cout << ans << endl;
}
		
	