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<bits/stdc++.h>
#define FLASH cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(false);
#define pb push_back
#define int long long
#define fr(i,a,b) for(int i=a;i<b;i++)
#define mod 1000000007
#define FILEIO freopen("/home/aman/Desktop/Kachda/input.txt", "r", stdin); freopen("/home/aman/Desktop/Kachda/output.txt", "w", stdout);

using namespace std;

int n, m, b = 0, w = 0;
char s[401][401];

bool vis[404][401];

void dfs(int i, int j){
	if(vis[i][j]) return;
	char clr = s[i][j];
	vis[i][j] = 1;
	if(clr == '.') b++;
	else w++;

	if(i > 1 and s[i - 1][j] != clr) dfs(i - 1, j);
	if(i < n and s[i + 1][j] != clr) dfs(i + 1, j);
	if(j > 1 and s[i][j - 1] != clr) dfs(i, j - 1);
	if(j < m and s[i][j + 1] != clr) dfs(i, j + 1); 
}
signed main()
{
	
 	FLASH


 	cin>>n>>m;
 	fr(i, 1, n + 1){
 		fr(j, 1, m + 1){
 			cin>>s[i][j];
 		}
 	}

 	int ans = 0;
 	
 	fr(i, 1, n + 1){
 		fr(j, 1, m + 1){
 			if(!vis[i][j]){
 				dfs(i,j);
 				ans += b*w;
 				b = w = 0;
 			}
 		}
 	}
 	cout<<ans;
}