G - Plus from Picture CodeForces - 1182B

Posted 2022-12-15 shallow-dream

You have a given picture with size w×hw×h. Determine if the given picture has a single "+" shape or not. A "+" shape is described below:

• A "+" shape has one center nonempty cell.
• There should be some (at least one) consecutive non-empty cells in each direction (left, right, up, down) from the center. In other words, there should be a ray in each direction.
• All other cells are empty.

Find out if the given picture has single "+" shape.

Input

The first line contains two integers hh and ww (1≤h1≤h, w≤500w≤500) — the height and width of the picture.

The ii-th of the next hh lines contains string sisi of length ww consisting "." and "*" where "." denotes the empty space and "*" denotes the non-empty space.

Output

If the given picture satisfies all conditions, print "YES". Otherwise, print "NO".

You can output each letter in any case (upper or lower).

Examples

Input

5 6
......
..*...
.****.
..*...
..*...

Output

YES

Input

3 5
..*..
****.
.*...

Output

NO

Input

7 7
.......
...*...
..****.
...*...
...*...
.......
.*.....

Output

NO

Input

5 6
..**..
..**..
******
..**..
..**..

Output

NO

Input

3 7
.*...*.
***.***
.*...*.

Output

NO

Input

5 10
..........
..*.......
.*.******.
..*.......
..........

Output

NO

Note

In the first example, the given picture contains one "+".

In the second example, two vertical branches are located in a different column.

In the third example, there is a dot outside of the shape.

In the fourth example, the width of the two vertical branches is 22.

In the fifth example, there are two shapes.

In the sixth example, there is an empty space inside of the shape.

 

 

#define TLE std::ios::sync_with_stdio(false);   cin.tie(NULL);   cout.tie(NULL);   cout.precision(10);
char ch[500+5][500+5];
int main()

    TLE;
    int n,m,x,y,flag=1;
    while(cin>>n>>m)
    
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            
                cin>>ch[i][j];
            

        int ans=0,xx=0;
        //debug(ch,n,m);
        int nx , ny;
        for(int i=0; i<n; i++)
        
            for(int j=0; j<m; j++)
            
                if(ch[i][j]==‘*‘ && ch[i][j-1]==‘*‘ && ch[i][j+1]==‘*‘ && ch[i-1][j]==‘*‘ && ch[i+1][j]==‘*‘  && j-1>=0 && j+1<m && i-1>=0 && i+1<n)
                
                    ch[i][j] = ‘.‘;
                    for(int kkkk=0; kkkk<4; kkkk++)
                    
                        nx = dir[kkkk][0] + i ;
                        ny = dir[kkkk][1] + j ;
                        while(ch[nx][ny]==‘*‘)
                        
                            ch[nx][ny] = ‘.‘;
                            nx += dir[kkkk][0];
                            ny += dir[kkkk][1];
                        
                    
                    xx=1;break;
                
            
            if(xx) break;
        

        int kk=0;
        for(int i=0; i<n; i++)
        
            for(int j=0; j<m; j++)
            
                if(ch[i][j]==‘*‘)
                
                    kk=1;
                    break;
                
            
            if(kk) break;
        
        if(xx==0 || kk )
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    
    ok;