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 PaintingPebbles {
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int k = cin.nextInt();
        int[] a = new int[n];
        int[] w = new int[n];
        for(int i = 0 ; i < n ; i ++)
        {
            a[i] = cin.nextInt();
            w[i] = a[i];
        }
        
        int[][] b = new int[n][k];
        for(int i = 0 ; i < n ; i ++)
        {
            for(int j = 0 ; j < k ; j ++)
                b[i][j] = 0;
            
            int flag = 0;
            while(a[i] > 0)
            {
                a[i] --;
                b[i][flag % k] ++;
                flag ++;
            }
        }
        
        if(!test(b , n , k))
        {
            System.out.println("NO");
        }
        else
        {
            System.out.println("YES");
            for(int i = 0 ; i < n ; i ++)
            {
                int flag = 0;
                for(int j = 0 ; j < k ; j ++)
                {
                    while(b[i][j] > 0)
                    {
                        b[i][j] --;
                        if(flag == 0)
                        {
                            System.out.print(j + 1);
                            flag = 1;
                        }
                        else
                            System.out.print(" " + (j + 1));
                    }
                }
                System.out.println();
            }
        }
        
    }
    
    public static boolean test(int[][] b , int n , int k)
    {
        boolean flag = true;
        
        for(int i = 0 ; i < k ; i ++)
            for(int j = 0 ; j < n ; j ++)
            {
                for(int l = 0 ; l < j ; l ++)
                {
                    if(Math.abs(b[j][i] - b[l][i]) > 1)
                        return false;
                }
            }
        
        
        return flag;
    }

}