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<cstdio>
using namespace std;

int visited[160001],used[160001];
long cnt,cnt2;
int h,w;
char s[410][410];
void dfs(int x,int y){
    visited[x*w+y]=1;
    used[x*w+y]=1;
    if(s[x][y]=='#'){
        cnt2++;
    }
    cnt++;
    if(x>0&&s[x][y]!=s[x-1][y]&&used[(x-1)*w+y]==0){
        dfs(x-1,y);
    }
    if(x<h-1&&s[x][y]!=s[x+1][y]&&used[(x+1)*w+y]==0){
        dfs(x+1,y);
    }
    if(y>0&&s[x][y]!=s[x][y-1]&&used[(x)*w+y-1]==0){
        dfs(x,y-1);
    }
    if(y<w-1&&s[x][y]!=s[x][y+1]&&used[(x)*w+y+1]==0){
        dfs(x,y+1);
    }
}

int main(){
    long ans=0;
    scanf("%d %d",&h,&w);
    for(int i=0;i<h;i++){
        scanf("%s",s[i]);
    }
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
            cnt=0;
            cnt2=0;
            if(used[i*w+j]==0){
                dfs(i,j);
                ans+=(cnt-cnt2)*cnt2;
            }
        }
    }
    printf("%ld\n",ans);
}