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;
using LL = long long;
using PII = pair<int, int>;

const int N = 410;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};

int h, w, cnt0, cnt1;
char s[N][N];
bool vis[N][N];
LL ans;

void dfs(int x, int y) {
    vis[x][y] = 1;
    if (s[x][y] == '.') cnt0++;
    else cnt1++;
    for (int i = 0; i < 4; i++) {
        int tx = x + dx[i], ty = y + dy[i];
        if (tx < 1 || tx > h || ty < 1 || ty > w) continue;
        if (!vis[tx][ty] && s[tx][ty] != s[x][y]) dfs(tx, ty);
    }

}

int main() {
    scanf("%d%d", &h, &w);
    for (int i = 1; i <= h; i++) scanf("%s", s[i] + 1);
    for (int i = 1; i <= h; i++)
        for (int j = 1; j <= w; j++) if (!vis[i][j] && s[i][j] == '#') {
            cnt0 = cnt1 = 0;
            dfs(i, j);
            ans += 1LL * cnt0 * cnt1;
        }
    printf("%lld\n", ans);
    return 0;
}
