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.*;
public class marte{


private int contor=0;

public static void main (String args[]) {

Scanner input=new Scanner(System.in);

HashMap<Integer, Integer> map1= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> map2= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> finish= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> maxime= new HashMap<Integer, Integer>();
	
			
			int n=input.nextInt();
			int k=input.nextInt();
			
			int nums[]=new int[n];
			int max=-1;
			int min=Integer.MAX_VALUE;
			
			
			for(int i=0;i<n;i++){
			nums[i]=input.nextInt();
			max=Math.max(max,nums[i]);
			min=Math.min(min,nums[i]);
			
			
			}
			
			if(max-min>k) System.out.println("NO");
			else{
			    System.out.println("YES");
			    
			for (int i=0;i<n;i++)
			{
			for(int j=1;j<=nums[i];j++) {
			    if(j%k==0) System.out.print(k+" ");
			     
		else 	    System.out.print((j)%(k)+" ");
			    
			    
			}
			System.out.println();
			
			
			}
			
			
			}
			
			
		
	}

    
		}