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

import java.util.Arrays;
import java.util.Scanner;

public class Solution 
{
    
     public static void main(String a[]) 
     {
         int n,k,c=1;
         Scanner sc=new Scanner(System.in); 
         n=sc.nextInt();
         k=sc.nextInt();
         
         int arr[]=new int[n];
         int temp[]=new int[n];
         
        
         
         for(int i=0;i<n;i++)
         {
             arr[i]=sc.nextInt();
             temp[i]=arr[i];
         }
         
         Arrays.sort(temp);
        if(temp[n-1]>((temp[0]*2)+(k-temp[0])))
        {
            System.out.println("NO");
        }
        else
        {
            System.out.println("YES");
            for(int j=0;j<n;j++)
             {
                 int ele=arr[j];
                 c=1;
                 for(int m=0;m<ele;m++)
                 {
                     if(c<=k)
                     {
                        System.out.print(c +" ");
                                 c++;
                     }
                     else
                     {
                         c=1;
                         System.out.print(c +" ");
                         c++;
                     }
                 }
                 System.out.println(" ");
             }
        }
         
         
         
         
     }
 
    
}
