Problem: 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

Solution (in CPP):

Solution:
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define MAXN 405
#define INF 1000000000
#define MOD 1000000007
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
ll H,W,ans,cnt0,cnt1;
char str[MAXN][MAXN];
bool vis[MAXN][MAXN];
void dfs(ll x,ll y)
{
    if(str[x][y]=='#') cnt0++; else cnt1++;
    vis[x][y]=true;
    if(x>1&&!vis[x-1][y]&&str[x-1][y]!=str[x][y]) dfs(x-1,y);
    if(x<H&&!vis[x+1][y]&&str[x+1][y]!=str[x][y]) dfs(x+1,y);
    if(y>1&&!vis[x][y-1]&&str[x][y-1]!=str[x][y]) dfs(x,y-1);
    if(y<W&&!vis[x][y+1]&&str[x][y+1]!=str[x][y]) dfs(x,y+1);
}
int main()
{
    scanf("%lld%lld",&H,&W);
    for(ll i=1;i<=H;i++) scanf("%s",str[i]+1);
    memset(vis,false,sizeof(vis));
    ans=0;
    for(ll i=1;i<=H;i++)
    {
        for(ll j=1;j<=W;j++)
        {
            if(vis[i][j]) continue;
            cnt0=cnt1=0;dfs(i,j);
            ans+=cnt1*cnt0;
        }
    }
    printf("%lld\n",ans);
    return 0;
}