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 mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
#define mkp make_pair
#define lP pair<ll, ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;
const int X[4] = { 0,0,1,-1 }, Y[4] = { 1,-1,0,0 };

int H, W, cnt[2];
ll res;
bool b[400][400];
char c[400][400];

void dfs(int x, int y) {
	b[x][y] = true;
	if (c[x][y] == '.')
		++cnt[0];
	else
		++cnt[1];
	for (int i = 0; i != 4; ++i) {
		int xx = x + X[i], yy = y + Y[i];
		if (xx == -1 || yy == -1 || xx == H || yy == W || b[xx][yy])continue;
		if (c[x][y] != c[xx][yy])
			dfs(xx, yy);
	}
}

int main() {
	cin >> H >> W;
	for (int i = 0; i != H; ++i) {
		for (int j = 0; j != W; ++j) {
			cin >> c[i][j];
		}
	}
	for (int i = 0; i != H; ++i) {
		for (int j = 0; j != W; ++j) {
			if (!b[i][j]) {
				dfs(i, j);
				res += (ll)cnt[0] * cnt[1];
				cnt[0] = cnt[1] = 0;
			}
		}
	}
	cout << res << endl;
	return 0;
}