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 <iostream>
using namespace std;
#define repl(i, l, r) for (int i = l; i < r; i++)
#define rep(i, n) repl(i, 0, n)
const int dx[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
const int dy[9] = {0, 1, 0, -1, 1, 1, -1, -1, 0};
char c[405][405];
bool seen[400][400];
int h, w;
long long dfs(const int &i, const int &j, long long &ca, long long &cb) {
	//連結成分内の黒の数をca，白の数をcb
	if (seen[i][j]) return 0;  // 1回見られていたら見なくてよい
	seen[i][j] = 1;
	if (c[i][j] == '#')  //今見ているマスが黒か白か
		ca++;
	else
		cb++;
	rep(k, 4) {  // 4回で下右上左の順に隣接マスを見ていく
		int nx = i + dx[k], ny = j + dy[k];
		if (nx < 0 or h <= nx or ny < 0 or w <= ny) continue;  //マスからはみ出していないか
		if (c[i][j] != c[nx][ny]) dfs(nx, ny, ca, cb);         //隣が違う色のマスなら探索
	}
	return ca * cb;
}
int main() {
	scanf("%d%d", &h, &w);
	rep(i, h) scanf("%s", c[i]);  // cが与えられた二次元配列
	long long ans = 0;                  //答え
	rep(i, h) rep(j, w) {
		long long ca = 0, cb = 0;
		ans += dfs(i, j, ca, cb);  //連結成分内の黒の数と白の数の積を答えに足していく
	}
	cout << ans << endl;
	return 0;
}