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.Scanner;


public class div289_2 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner input=new Scanner(System.in);
		int n=input.nextInt();
		int k=input.nextInt();
		
		int[] p=new int[n+1];
		int[][] pl=new int[n+1][101];
		
		int max=0;
		for(int i=1;i<=n;i++)
		{
			int t=input.nextInt();
			p[i]=t;
			if(t>max)max=t;
			
			for(int j=1;j<=t;j++)pl[i][j]=0;
		}
		int c=1,change_color=0,ck=0,prev=n,f=0;
		
		for(int i=1;i<=max;i++)
		{
			change_color=0;
			ck=0;
			for(int j=1;j<=n;j++)
			{
				 if(i==1){
					 pl[j][i]=c;
					 
				 }
				 else if(i!=1){
					 
					 if(i<=p[j]){
						 pl[j][i]=c;
						 ck++;
						 
					 }
					 else
					 {
						 if(i-1<=p[j]){
							 change_color=1;
						 }
					 }
				 }
			 }	
			if(change_color==1){
				c++;
			}
			if(change_color==0 && ck!=n){
				if(ck==prev){
					c++;
					
				}
			}
			if(c>k && i!=max){
					{
					f=1;break;
					}
			}
			prev=ck;
				
			
		}
		if(f==0){
			System.out.print("YES\n");
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=p[i];j++){
					System.out.print(pl[i][j]+" ");
				}
				System.out.print("\n");
			}
			
		}
		else
		{
			System.out.print("NO");
		}

	}

}
