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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <climits>
#include <clocale>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i, s, n) for(int i = (s), i##_len=(n); i < i##_len; ++i)
#define FORS(i, s, n) for(int i = (s), i##_len=(n); i <= i##_len; ++i)
#define VFOR(i, s, n) for(int i = (s); i < (n); ++i)
#define VFORS(i, s, n) for(int i = (s); i <= (n); ++i)
#define REP(i, n) FOR(i, 0, n)
#define REPS(i, n) FORS(i, 0, n)
#define VREP(i, n) VFOR(i, 0, n)
#define VREPS(i, n) VFORS(i, 0, n)
#define RFOR(i, s, n) for(int i = (s), i##_len=(n); i >= i##_len; --i)
#define RFORS(i, s, n) for(int i = (s), i##_len=(n); i > i##_len; --i)
#define RREP(i, n) RFOR(i, n, 0)
#define RREPS(i, n) RFORS(i, n, 0)
#define IREP(i, v) for(auto i = (v).begin(); i != (v).end(); ++i)
#define IRREP(i, v) for(auto i = (v).rbegin(); i != (v).rend(); ++i)
#define ALL(v) (v).begin(), (v).end()
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v), greater<decltype(v[0])>())
#define SZ(x) ((int)(x).size())
#define REV(x) reverse(ALL(x))
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define MT make_tuple
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(ALL(v)), v.end())

using ld = long double;
using ll = long long;
using ui = unsigned int;
using ull = unsigned long long;
using Pi_i = pair<int, int>;
using Pll_ll = pair<ll, ll>;
using VB = vector<bool>;
using VC = vector<char>;
using VD = vector<double>;
using VI = vector<int>;
using VLL = vector<ll>;
using VS = vector<string>;
using VSH = vector<short>;
using VULL = vector<ull>;

const int MOD = 1000000007; 
const int INF = 1000000000; //1e9
const int NIL = -1;
const ll LINF = 1000000000000000000; // 1e18
const double EPS = 1E-10;

template<class T, class S>
bool chmax(T &a, const S &b){
    if(a < b){
        a = b; return true;
    }
    return false;
}
template<class T, class S>
bool chmin(T &a, const S &b){
    if(b < a){
        a = b; return true;
    }
    return false;
}


int main(){
    int H, W; cin >> H >> W;
    vector<VB> Sq(H, VB(W));
    vector<VI> kind(H, VI(W, -1));
    REP(i, H){
        string S; cin >> S;
        REP(j, W)
            Sq[i][j] = (S[j] == '#');
    }
    int id = -1;
    REP(i, H){
        REP(j, W){
            if(kind[i][j] != -1) continue;
            kind[i][j] = ++id;
            queue<Pi_i> que;
            que.push({i, j});
            while(!que.empty()){
                int dx[4] = {0, -1, 0, 1};
                int dy[4] = {1, 0, -1, 0};
                int ii(que.front().first), jj(que.front().second);
                que.pop();
                bool blk(Sq[ii][jj]);
                REP(a, 4){
                    int ni(ii + dx[a]), nj(jj + dy[a]);
                    if(ni < 0 || H <= ni || nj < 0 || W <= nj || kind[ni][nj] != -1)
                        continue;
                    if(Sq[ni][nj] == blk) continue;
                    kind[ni][nj] = id;
                    que.push({ni, nj});
                }
            }
        }
    }
    ++id;
    VI cntB(id, 0), cntW(id, 0);
    REP(i, H){
        REP(j, W){
            if(Sq[i][j]) ++cntB[kind[i][j]];
            else ++cntW[kind[i][j]];
        }
    }
    ll ans(0);
    REP(i, id){
        ans += (ll)cntB[i] * cntW[i];
    }
    cout << ans << endl;
    return 0;
}