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 ll  = long long;
#define int ll
#define rep(i,n) for(int i=0;i<n;i++)
#define all(in) in.begin(), in.end()
using namespace std;
int W,H;
int dx[]={0,0,1,-1}, dy[]={1,-1,0,0};
bool valid(int x,int y){return (0<=x&&x<W)&&(0<=y&&y<H);}
vector<string>g;
vector<vector<bool>>used;
int cnt = 0;
int dfs(int y,int x){
    used[y][x] = true;
    char c = g[y][x];
    if(c == '#')cnt++;
    int cal = (c == '.');
    rep(k,4){
        int ny = y + dy[k];
        int nx = x + dx[k];
        if(valid(nx, ny) and c != g[ny][nx] and !used[ny][nx]){
            cal += dfs(ny,nx);
        }
    }
    return cal;
}
signed main(){
    cin >> H >> W;
    g = vector<string>(H);
    used = vector<vector<bool>>(H,vector<bool>(W,false));
    for(auto & s :g) cin >> s;
    int ans = 0;
    rep(i,H)rep(j,W)if(g[i][j] == '#' and !used[i][j]){
        ans += dfs(i,j) * cnt;
        cnt = 0;
    }
    cout << ans << endl;
}


