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>
using namespace std;
#define LL long long
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};
char mp[405][405];
//bool vis[405][405];
int belong[405][405];
pair<int, int> S[160010];
int Scnt;
int h, w;
void dfs(int x, int y, int s)
{
	belong[x][y] = s;
	if (mp[x][y] == '#')
		++S[s].first;
	else
		++S[s].second;
	for (int i = 0; i < 4; ++i)
	{
		int nx = x + dx[i], ny = y + dy[i];
		if (nx > 0 && nx <= h && ny > 0 && ny <= w && mp[x][y] != mp[nx][ny] && belong[nx][ny] == 0)
		{
			dfs(nx, ny, s);
		}
	}
}
void input()
{
	cin >> h >> w;
	for (int i = 1; i <= h; ++i)
	{
		cin >> (mp[i] + 1);
	}
}

void solve()
{
	for (int i = 1; i <= h; ++i)
		for (int j = 1; j <= w; ++j)
			if (!belong[i][j])
				dfs(i, j, ++Scnt);
	long long ans = 0;
	for (int i = 1; i <= Scnt; ++i)
		ans += 1LL * S[i].first * S[i].second;
	cout << ans;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	input();
	solve();
	return 0;
}
