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 <cstdio>
#include <cstring>
#include <utility>
#define ll long long
#define P pair<ll, ll>
using namespace std;

int H, W;
char S[401][400];

bool used[400][400];
int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 };

P dfs(int h, int w) {
	P res;
	if (S[h][w] == '#') res.first++;
	else res.second++;
	used[h][w] = true;
	for (int i = 0; i < 4; i++) {
		int nx = h + dx[i], ny = w + dy[i];
		if (nx >= 0 && nx < H && ny >= 0 && ny < W && !used[nx][ny] && S[h][w] != S[nx][ny]) {
			P p = dfs(nx, ny);
			res.first += p.first;
			res.second += p.second;
		}
	}
	return res;
}

void solve() {
	memset(used, false, sizeof(used));
	ll ans = 0;
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (!used[i][j]) {
				P res = dfs(i, j);
				ans += res.first * res.second;
			}
		}
	}
	printf("%lld\n", ans);
}

int main() {
	scanf("%d %d", &H, &W);
	for (int i = 0; i < H; i++)
		scanf("%s", S[i]);
	solve();
}