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<iostream>
#include<bitset>
#include<string.h>
using namespace std;
typedef long long ll;
bool vis[401][401];
char mat[401][401];
int H,W;
ll blk, wht;
void dfs(int i, int j){
    vis[i][j] = true;
    int r[] = {1, -1, 0, 0}, c[] = {0, 0, 1, -1};
    int cnt = 0;
    if (mat[i][j] == '#') blk++;
    if (mat[i][j] == '.') wht++;
    for (int x=0;x<4;x++){
        int ni = i + r[x], nj = j + c[x];
        if (ni < H && ni >=0 && nj < W && nj >= 0 && !vis[ni][nj] && mat[ni][nj]!= mat[i][j])
            dfs(ni, nj);
    }
    // cout << "cnt: " << cnt << endl;
}   

int main(void){
    cin >> H >> W;
    for (int i=0;i<H;i++)
        cin >> mat[i];
    memset(vis,false,sizeof(vis));
    ll ans = 0;
    for (int i=0;i<H;i++){
        for (int j=0;j<W;j++){
            if (vis[i][j]) continue;
            blk = 0;
            wht = 0;
            dfs(i, j);
            ans += blk*wht;
        }
    }
    cout << ans << endl;
}