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<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<deque>
using namespace std;
#define ll long long
const int mod = 1000000007;
const ll INF = 1000000000000000000;

int H, W;
string S[410];
bool visit[410][410];
int vx[] = { 0,1,0,-1 };
int vy[] = { 1,0,-1,0 };
ll white, black;
void dfs(int x, int y) {
	visit[y][x] = true;
	if (S[y][x] == '.') white++;
	else black++;
	for (int i = 0; i < 4; i++) {
		int nx = x + vx[i];
		int ny = y + vy[i];
		if (nx < 0 || nx >= W || ny < 0 || ny >= H) continue;
		if (visit[ny][nx]) continue;
		if (S[y][x] == S[ny][nx]) continue;
		dfs(nx, ny);
	}
}
int main()
{
	cin >> H >> W;
	for (int i = 0; i < H; i++) cin >> S[i];
	ll ans = 0;
	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (visit[i][j]) continue;
			//cout << i << " " << j << endl;
			white = 0; black = 0;
			dfs(j, i);
			ans += white * black;
		}
	}
	cout << ans << endl;
}
