Problem: There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Examples

Input

4 4
1 2 3 4


Output

YES
1
1 4
1 2 4
1 2 3 4


Input

5 2
3 2 4 1 3


Output

NO


Input

5 4
3 2 4 3 5


Output

YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

Solution (in JAVA):

Solution:
import java.util.Scanner;
public class paintingPebbles {

    static int update[] = new int[1001];

    public static void main(String args[]) {

        Scanner sc = new Scanner(System.in);
        int n =sc.nextInt();
        int  k =sc.nextInt();
        int a[]=new int[n];
        int max=Integer.MIN_VALUE;
        int min=Integer.MAX_VALUE;
       
        for(int i=0;i<n;i++)
        {
            a[i]=sc.nextInt();
            if(a[i]>max) max=a[i];
            
            if(a[i]<min) min=a[i];
            
                
        }
         
        if(k>=max)
        {
          System.out.println("YES");
          for(int i=0;i<n;i++)
          {
              for(int j=1;j<=a[i];j++) System.out.print(j+" ");
                  
              
              System.out.println();
          }
            
        
        }
        else
        {
           int t=max/k;
           
           if(max>=t*k && min>=(t-1)*k + max%k)
           {
               System.out.println("YES");
               for(int i=0;i<n;i++)
          {
             int c=1;
              for(int j=1;j<=a[i];j++) 
              {
                  
                 
                  System.out.print(c+" ");
                  if(c==k)
                      c=1;
                  else c++;
                  
                  
              }
              System.out.println();
               
           }
        
        
        
        }
           else System.out.println("NO");
        


    }
        

}
    
}