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;
typedef long long ll;
inline ll _() {
	ll x=0,f=1; char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())
		if(ch=='-')f=-f;
	for(;ch>='0'&&ch<='9';ch=getchar())
		x=x*10+ch-'0';
	return x*f;
}
#define _ _()
int n,m;
ll f[405*405],w[405*405],b[405*405];
ll ans;
char s[405][405];
const int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
inline int find( int x ) { return f[x]==x?x:f[x]=find(f[x]); } 
int main() {
	n=_; m=_;
	for( int i=1;i<=n*m;i++ ) f[i]=i;
	for( int i=1;i<=n;i++ ) scanf("%s",s[i]+1);
	for( int i=1;i<=n;i++ ) 
		for( int j=1;j<=m;j++ ) {
			for( int k=0;k<4;k++ ) {
				int nx=i+dx[k],ny=j+dy[k];
				if(nx<1||ny<1||nx>n||ny>m) continue;
				if(s[nx][ny]!=s[i][j]) {
					int fx=find((nx-1)*m+ny),fy=find((i-1)*m+j);
					f[fx]=f[fy]; 
				}
			}
		}
	for( int i=1;i<=n;i++ ) 
		for( int j=1;j<=m;j++ ) 
			if(s[i][j]=='#') b[find((i-1)*m+j)]++;
			else w[find((i-1)*m+j)]++;
	for( int i=1;i<=n*m;i++ )
		ans+=b[i]*w[i];
	printf("%lld\n",ans);
}