Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int MAX_HW = 405;
char c[MAX_HW][MAX_HW];
vector<int> B;
bool visited[MAX_HW][MAX_HW];
int dx[4] = {0,1,0,-1}, dy[4] = {1,0,-1,0};
ll cntB = 0, cntW = 0;
int H,W;

void dfs(int x, int y)
{
	if (visited[x][y]) return;
	visited[x][y] = true;
	if (c[x][y] == '#') cntB++;
	else cntW++;
	rep(i,4)
	{
		int xx = x+dx[i], yy = y+dy[i];
		if (xx >= 0 && xx < H && yy >= 0 && yy < W)
		{
			if (c[x][y] == '#' && c[xx][yy] == '.')
			{
				dfs(xx,yy);
			}
			else if (c[x][y] == '.' && c[xx][yy] == '#')
			{
				dfs(xx,yy);
			}
		}
	}
	return;
}

int main()
{
	cin >> H >> W;
	rep(i,H)
	{
		rep(j,W)
		{
			cin >> c[i][j];
		}
	}

	ll ans = 0;
	rep(i,H)
	{
		rep(j,W)
		{
			cntB = 0, cntW = 0;
			dfs(i,j);
			ans += cntB * cntW;
		}
	}
	cout << ans << endl;
	return 0;
}