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;
using ull = unsigned long long;
using P = pair<int,int>;
#define rep(i,n) for(ll i = 0;i < (ll)n;i++)
#define ALL(x) (x).begin(),(x).end()
#define MOD 1000000007

int dx[] = {1,0,0,-1};
int dy[] = {0,1,-1,0};
int h,w;
vector<vector<bool>> seen(500,vector<bool>(500,0));
vector<string> s(500);
pair<ll,ll> dfs(int y,int x){
  pair<ll,ll> res = make_pair((s[y][x] == '#'),(s[y][x] != '#'));
  seen[y][x] = true;
  rep(i,4){
    int ny = y+dy[i],nx = x+dx[i];
    if(ny < 0 || ny >= h || nx < 0 || nx >= w)continue;
    if(s[ny][nx] == s[y][x])continue;
    if(seen[ny][nx])continue;
    pair<ll,ll> k = dfs(ny,nx);
    res.first += k.first,res.second += k.second;
  }
  return res;
}

int main(){
  
  
  cin >> h >> w;
  rep(i,h)cin >> s[i];
  ll res = 0;
  rep(i,h)rep(j,w){
    if(s[i][j] != '#')continue;
    if(seen[i][j])continue;
    pair<ll,ll> k = dfs(i,j);
    res += k.first*k.second;
  }
  cout << res << endl;








  return 0;
}