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 int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
 
string G[405];
bool seen[405][405];
ll H, W, cnt_w, cnt_b;
 
void dfs(int h, int w) {
    seen[h][w] = true;
    if (G[h][w] == '.') cnt_w++;
    else cnt_b++;
 
    for (int dir = 0; dir < 4; ++dir) {
        int nh = h + dx[dir];
        int nw = w + dy[dir];
 
        if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
        if (G[nh][nw] == G[h][w]) continue;
 
        if (seen[nh][nw]) continue;
 
        dfs(nh, nw);
    }
 
    return;
}
 
int main() {
    ll ans = 0;
    cin >> H >> W;
    for (int h = 0; h < H; h++) cin >> G[h];
 
    for (int h = 0; h < H; h++) {
        for (int w = 0; w < W; w++) {
            if (!seen[h][w]) {
                cnt_b = 0;
                cnt_w = 0;
                dfs(h, w);
                ans += cnt_b * cnt_w;
            }
        }
    }
    cout << ans << endl;
    return 0;
}