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 mp make_pair
#define pb push_back

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
const int INF = 987654321;
const int MOD = 1000000007;

int H,W;
char mat[401][401];
bool chk[401][401];
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
ll ans = 0;
ll tmp = 0, cnt = 1;

void dfs(int sy, int sx, int idx){
  chk[sy][sx] = true;
  for(int i=0;i<4;i++){
    int y = sy+dir[i][0], x = sx+dir[i][1];
    if(y < 0 || y >= H || x < 0 || x >= W) continue;
    if(!chk[y][x] && mat[y][x] == (idx == 0 ? '.' : '#')){
      if(mat[y][x] == '.') tmp++;
      else cnt++;
      dfs(y,x,!idx);
    }
  }
}

int main(){
  scanf("%d %d",&H,&W);
  for(int i=0;i<H;i++)
    scanf("%s",mat[i]);
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      if(mat[i][j] == '#' && chk[i][j] == false){
        tmp = 0; cnt = 1;
        dfs(i,j,0);
        ans += cnt * tmp;
      }
    }
  }
  printf("%lld\n",ans);
}
