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 <utility>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
int H,W;
char F[410][410];
int visited[410][410] = {0};
int dx[4] = {0,0,1,-1},dy[4] = {1,-1,0,0};

bool in(int x,int y){
	return 1<=x && x<=H && 1<=y && y<=W;
}

pair<ll,ll> dfs(int x,int y,int c){
	visited[x][y] = 1;
	P cnt;
	if(c==0) cnt = {1,0};
	else cnt = {0,1};
	for(int i=0;i<4;i++){
		int nx = x+dx[i],ny = y+dy[i];
		if(visited[nx][ny]==0 && in(nx,ny)){
			if(c==0 && F[nx][ny]=='.'){
				P p = dfs(nx,ny,1);
				cnt.first += p.first;
				cnt.second += p.second;
			}
			else if(c==1 && F[nx][ny]=='#'){
				pair<ll,ll> p = dfs(nx,ny,0);
				cnt.first += p.first;
				cnt.second += p.second;
			}
		}
	}
	return cnt;
}

int main(){
	cin >> H >> W;
	for(int i=1;i<=H;i++){
		for(int j=1;j<=W;j++){
			cin >> F[i][j];
		}
	}
	ll ans = 0;
	for(int i=1;i<=H;i++){
		for(int j=1;j<=W;j++){
			if(F[i][j]=='#' && visited[i][j]==0){
				P t = dfs(i,j,0);
				ans += t.first*t.second;
			}
		}
	}
	cout << ans << endl;
}