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;
typedef pair<ll,ll> P;

ll H,W;
string s[400];
ll m[400][400];
ll dh[4] = {0,1,0,-1};
ll dw[4] = {1,0,-1,0};

P dfs(ll h, ll w){
    ll A = 0;
    ll B = 0;
    m[h][w] = 1;
    for(int i=0;i<4;i++){
        ll nh = h + dh[i];
        ll nw = w + dw[i];
        if(nh < 0 || nh >= H) continue;
        if(nw < 0 || nw >= W) continue;
        if(s[h][w] == s[nh][nw]) continue;
        if(m[nh][nw] != 0) continue;
        P p = dfs(nh, nw);
        A += p.first;
        B += p.second;
    }
    if(s[h][w] == '#') A++;
    else B++;
    return {A,B};
}

int main(){
    cin >> H >> W;
    for(int i=0;i<H;i++) cin >> s[i];
    ll ans = 0;
    for(int h=0;h<H;h++){
        for(int w=0;w<W;w++){
            if(m[h][w] == 0){
                P p = dfs(h,w);
                ans += p.first * p.second;
            }
        }
    }
    cout << ans << endl;
    return 0;
}