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 = long long;
#define rep(i, m, n) for (int i = m; i < n; ++i)
const int MOD = 1000000007;

int vx[4] = {1,0,-1,0};
int vy[4] = {0,1,0,-1};
using P = pair<ll, ll>;

int H, W;
vector<string> S;
vector<vector<bool>> visited;

P dfs(int y, int x) {
  P ret = {0, 0};
  if(S[y][x] == '.') ret.first++;
  else ret.second++;
  visited[y][x] = true;
  rep(i, 0, 4) {
    int nx = x + vx[i];
    int ny = y + vy[i];
    if(nx < 0 || W <= nx || ny < 0 || H <= ny) continue;
    if(visited[ny][nx]) continue;
    if(S[y][x] == S[ny][nx]) continue;
    P T = dfs(ny, nx);
    ret.first += T.first;
    ret.second += T.second;
  }
  return ret;
}

int main() {
  cin >> H >> W;
  S.resize(H);
  visited.assign(H, vector<bool>(W));
  rep(i, 0, H) cin >> S[i];
  ll ans = 0;
  rep(i, 0, H) rep(j, 0, W) {
    if(!visited[i][j]) {
      P tmp = dfs(i, j);
      ans += tmp.first * tmp.second;
    }
  }
  cout << ans << endl;
  return 0;
}
