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 pebbles1b18
{
	
public static void main(String args[])
{
	Scanner s=new Scanner(System.in);
	int n=s.nextInt();
	int k=s.nextInt();
	int[] a=new int[n];
	for(int i=0;i<n;i++)
	{
		a[i]=s.nextInt();
	}
	int min=a[0];
	for(int i=1;i<n;i++)
	{
		if(a[i]<min)
			min=a[i];
	}
	int max=a[0];
	for(int i=0;i<n;i++)
	{

		if(a[i]>max)
			max=a[i];
	}
	if(max-min>k)
	System.out.println("NO");
	else
	{
	System.out.println("YES");
	for(int u=0;u<n;u++)
	{
	int q=a[u]/k;
	int r=a[u]%k;
	
	for(int i=1;i<=r;i++)
	{
		for(int j=0;j<=q;j++)
		{
			System.out.print(i + " ");
		}
	}
	for(int i=r+1;i<=k;i++)
	{
		for(int j=0;j<q;j++)
		{
			System.out.print(i + " ");
		}
	}
	System.out.println();
	}
	}
}

}