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 REP(i,a,b) for(int i=(a);i<(b);i++)
#define rep(i,n) REP(i,0,(n))
typedef long long ll;
using namespace std;
     
int h, w, dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
string s[400];
bool visited[400][400];
ll ans, bk, wt;

void dfs(int sy, int sx, bool isWhite) {
    visited[sy][sx] = true;
    if (isWhite) wt++;
    else bk++;
    
    rep(i, 4) {   
        int ny = sy + dy[i], nx = sx + dx[i];
        if (ny < 0 || ny >= h || nx < 0 || nx >= w || visited[ny][nx]) continue;
        if ( isWhite && s[ny][nx] == '#') dfs(ny, nx, false);
        if (!isWhite && s[ny][nx] == '.') dfs(ny, nx, true);
    }
}
     
int main() {
    cin >> h >> w;
    rep(i, h) cin >> s[i];
    REP(i, 0, h) {
        REP(j, 0, w) {
            if (s[i][j] == '#') {
                bk = wt = 0;
                dfs(i, j, false);
                ans += bk * wt;
            }
        }
    }
    cout << ans << endl;
    return 0;
}
