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 = int64_t;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
vector<string> S;
ll H, W;
bool used[400][400];
ll dh[] = {1, 0, -1, 0};
ll dw[] = {0, 1, 0, -1};
P dfs(ll h, ll w){
    used[h][w] = 1;
    P ret = P(S[h][w] == '.', S[h][w] == '#');
    for(ll i = 0; i < 4; i++){
        ll nh = h + dh[i];
        ll nw = w + dw[i];
        if(!(0 <= nh && nh < H &&
             0 <= nw && nw < W)) continue;
        if(S[nh][nw] == S[h][w]) continue;
        if(used[nh][nw]) continue;
        P tmp = dfs(nh, nw);
        ret.first += tmp.first;
        ret.second += tmp.second;
    }
    return ret;
}

int main(){
    cin >> H >> W;
    S.resize(H);
    for(auto &s : S) cin >> s;
    ll ans = 0;
    for(ll i = 0; i < H; i++) for(ll j = 0; j < W; j++){
        if(used[i][j]) continue;
        auto tmp = dfs(i, j);
        ans += tmp.first * tmp.second;
    }
    cout << ans << endl;
    return 0;
}
