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;
#define rep(i,n) for(int i=0;i<int(n);++i)
using ll = long long;

string s[410];
int Y, X;

int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};

void calc(int y, int x, ll &black, ll &white, char pre) {
    if (y < 0 || x < 0 || x>=X || y>=Y) return;
    if (s[y][x] == '!' || s[y][x] == pre) return;
    if (s[y][x] == '#') black++;
    if (s[y][x] == '.') white++;
    char cur = s[y][x];
    s[y][x] = '!';
    rep(i,4){
        calc(y+dy[i], x+dx[i], black, white, cur);
    }
}

int main(void){
    cin >> Y >> X;
    rep(y,Y) cin >> s[y];

    ll res = 0;
    rep(y,Y) rep(x,X){
        ll black = 0;
        ll white = 0;
        calc(y, x, black, white, '!');
        res += black * white;
    }
    cout << res << endl;
    return 0;
}
