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;

using ll = long long;
using Pi = pair < int, int >;

constexpr int INF = (1 << 30) - 1;
constexpr int dy[] = { 1, 0, -1, 0, 1, -1, -1, 1 };
constexpr int dx[] = { 0, 1, 0, -1, 1, 1, -1, -1 };

int H, W;
string S[405];
ll ans = 0;

Pi dfs ( int y, int x )
{
	Pi ret = Pi(0, 0);
	char save = S[y][x];

	if ( S[y][x] == '#' ) ret.first++;
	else ret.second++;

	S[y][x] = 'N';

	for ( int i = 0; i < 4; i++ ) {
		int ny = y + dy[i], nx = x + dx[i];	
		if ( 0 <= ny && 0 <= nx && ny < H && nx < W && S[ny][nx] != 'N' ) {
			if ( save != S[ny][nx] ) {
				auto data = dfs(ny, nx);
				ret.first += data.first;
				ret.second += data.second;
			}
		}
	}

	return ( ret );
}

int main()
{
	cin >> H >> W;

	for ( int i = 0; i < H; i++ ) {
		cin >> S[i];
	}

	for ( int i = 0; i < H; i++ ) {
		for ( int j = 0; j < W; j++ ) {
			if ( S[i][j] != 'N' ) {
				auto data = dfs(i, j);
				ans += (ll)data.first * data.second;
			}
		}
	}
	
	cout << ans << endl;

	return ( 0 );
}
