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 static java.lang.Math.min;
import java.util.Scanner;

/**
 *
 * @author ESC
 */
public class JavaApplication35 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
           int k,n,mn=1000;
        boolean check=true ;
        
        
        int []arr= new int[100];
        Scanner in = new Scanner (System.in);
        n= in.nextInt(); k =in.nextInt();
        for(int i = 0 ; i < n ; i ++)
        {arr[i]=in.nextInt(); mn=min(arr[i],mn);}
        
        for(int i=0; i<n; i ++)
        { boolean c=false; if(arr[i]-mn<=k)c=true;
            check=(check&&c);
        
        }
        if(check==false){System.out.println("NO");}
        else {
            System.out.println("YES");
            for(int i=0; i <n; i++)
            {{for(int j = 0 ; j <arr[i];j++)
                System.out.print((j%k)+1+" ");
                }
            System.out.println();
            }
        }
        
    }
    
}