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;
typedef long long ll;
const ll MOD = 1e9 + 7;
const double PI = 3.141592653589793238;
const double EPS = 1e-10;
typedef pair<ll, ll> P;
bool vis[400][400];
string S[400];
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
int H, W;
P dfs(int x, int y) {
	vis[y][x] = 1;
	P ret(0, 0);
	if (S[y][x] == '#') ret.first++;
	else ret.second++;
	for (int i = 0; i < 4; i++) {
		int nx = x + dx[i], ny = y + dy[i];
		if (0 > nx || nx >= W || 0 > ny || ny >= H || vis[ny][nx]) continue;
		if (S[y][x] == S[ny][nx]) continue;
		P p = dfs(nx, ny);
		ret.first += p.first;
		ret.second += p.second;
	}
	return ret;
}
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 (vis[i][j]) continue;
			P p = dfs(j, i);
			ans += p.first*p.second;
		}
	}
	cout << ans << endl;
}