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
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"
#define rep(i,a,b) for (int i = (a); i < (b); i++)
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef complex<double> com;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const int inf = 2e9;

bool used[401][401];
int blk, wht, h, w;
vector<string> grid;
int dx[4] = { 1,-1,0,0 }, dy[4] = { 0,0,1,-1 };
void dfs(int x, int y) {
	used[x][y] = true;
	if (grid[x][y] == '#') blk++; else wht++;
	rep(i, 0, 4) {
		int tx = x + dx[i], ty = y + dy[i];
		if (tx < 0 || tx >= h || ty < 0 || ty >= w || used[tx][ty] || grid[x][y] == grid[tx][ty]) continue;
		dfs(tx, ty);
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> h >> w;
	grid.resize(h);
	rep(i, 0, h) cin >> grid[i];
	ll ans = 0;
	rep(i, 0, h)rep(j, 0, w) {
		if (grid[i][j] == '.'||used[i][j]) continue;
		blk = 0, wht = 0; dfs(i, j);
		ans += 1LL * blk * wht;
	}
	printf("%lld", ans);
	return 0;
}
