Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

3 3
.#.
..#
#..

Sample Output 1

10

Some of the pairs satisfying the condition are ((1, 2), (3, 3)) and ((3, 1), (3, 2)), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

Sample Input 2

2 4
....
....

Sample Output 2

0

Sample Input 3

4 3
###
###
...
###

Sample Output 3

6

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
constexpr long long INF = 1001001001LL;
constexpr long long LINF = 1e18 + 10;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define rept(i, j, n) for(int i=(j); i<(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

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

int main(){
    int H, W; cin >> H >> W;
    vector<string> grid(H);
    rep(i, H) cin >> grid[i];

    ll res = 0;
    rep(i, H){
        rep(j, W){
            //BFS 始点が白の場合のみ探索する
            if(grid[i][j]=='#')continue;
            queue<pii> que; //(y, x)
            que.push({i, j});
            vector<vector<int>> d(H, vector<int>(W)); 
            d[i][j] = 0;
            while(!que.empty()){
                auto p = que.front(); que.pop();
                int h = p.first, w = p.second;
                rep(i, 4){
                    int ny = h + dy[i];
                    int nx = w + dx[i];
                    int cost = d[h][w] + 1;
                    if(ny<0 or nx<0 or ny>=H or nx>=W)continue;
                    if(d[ny][nx] != 0)continue;
                    if(grid[ny][nx]=='.' and cost%2==0 or grid[ny][nx]=='#' and cost%2==1){
                        d[ny][nx] = cost;
                        que.push({ny, nx});
                        if(cost%2)res++;                                       
                    }
                    
                }
            }   
        }
    }
    cout << res << ln;
}