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"
#define in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

LL H, W;
std::vector<std::string>S;
std::vector<std::vector<bool>>vis;
std::vector<std::pair<LL, LL>>memo;

LL dx[] = { -1,0,0,1 }, dy[] = { 0,-1,1,0 };
void f(LL i, LL j)
{
	if (vis[i][j]) return;
	vis[i][j] = true;
	memo.push_back(std::make_pair(i, j));
	rep(k, 4)
	{
		LL x = i + dx[k], y = j + dy[k];
		if (x < 0 || H <= x || y < 0 || W <= y) continue;
		if (S[x][y] != S[i][j]) f(x, y);
	}
}

int main()
{
	in >> H >> W;
	S.resize(H);
	rep(i, H) in >> S[i];

	vis.resize(H);
	rep(i, H) vis[i].resize(W, false);

	LL ans = 0;
	rep(i, H)
	{
		rep(j, W)
		{
			if (!vis[i][j])
			{
				memo.clear();
				f(i, j);
				LL cnt1 = 0, cnt2 = 0;
				for (auto k : memo)
				{
					if (S[k.first][k.second] == '#') ++cnt1;
					else ++cnt2;
				}
				ans += cnt1 * cnt2;
			}
		}
	}

	out << ans << std::endl;
}
