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>
#define REP(i, a, n) for (ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

ll H, W;
char S[500][500];
bool visited[500][500];
ll dd[5] = { -1, 0, 1, 0, -1 };

void dfs(ll y, ll x, ll &b, ll &w) {
  if (visited[y][x]) return;
  visited[y][x] = true;

  if (S[y][x] == '#') b++;
  else w++;

  REP(i, 0, 4) {
    ll ny = y + dd[i], nx = x + dd[i + 1];
    if (0 <= ny && ny < H && 0 <= nx && nx < W && S[y][x] != S[ny][nx]) {
      dfs(ny, nx, b, w);
    }
  }
}

int main(void) {
  cin >> H >> W;
  REP(i, 0, H) cin >> S[i];

  ll ans = 0;
  REP(i, 0, H) REP(j, 0, W) {
    if (!visited[i][j]) {
      ll b = 0, w = 0;
      dfs(i, j, b, w);
      ans += b * w;
    }
  }
  cout << ans << endl;
}