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 main
{
    public static void main(String args[])
    {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int k = input.nextInt();
        int arr[] = new int[n];
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        for(int i=0;i<n;i++)
        {
            arr[i] = input.nextInt();
            if(arr[i] > max)
            {
                max = arr[i];
            }
            if(arr[i] < min)
            {
                min = arr[i];
            }
        }
        if(max - min <= k)
        {
            System.out.println("YES");
        }
        else
        {
            System.out.println("NO");
            System.exit(0);
        }
        for(int f=0;f<n;f++)
        {
            int a = arr[f]/k;
            int b = arr[f]%k;
            for(int i=1;i<=b;i++)
            {
                for(int j=1;j<=a+1;j++)
                {
                    System.out.print(i+" ");
                }
            }
            for(int i=b+1;i<=k;i++)
            {
                for(int j=1;j<=a;j++)
                {
                    System.out.print(i+" ");
                }
            }
            System.out.println();
        }
    }   
}