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 b {
public static void main(String[] args)
{
    Scanner input = new Scanner(System.in);
    int n = input.nextInt(), k = input.nextInt();
    int[] as = new int[n];
    int[][] res = new int[n][];
    int min = 101;
    for(int i = 0; i<n; i++)
    {
        as[i] = input.nextInt();
        min = Math.min(as[i], min);
        res[i] = new int[as[i]];
    }
    int[] ptrs = new int[n];
    for(int i = 0; i<n; i++)
        for(int j = 0; j<min+1 && j < as[i]; j++)
        {
            res[i][ptrs[i]] = 1;
            ptrs[i]++;
        }
    for(int i = 2; i<=k; i++)
    {
        for(int j = 0; j<n; j++)
        {
            if(ptrs[j] < as[j])
            {
                res[j][ptrs[j]] = i;
                ptrs[j]++;
            }
        }
    }
    boolean good = true;
    for(int i = 0; i<n; i++) good &= ptrs[i] == as[i];
    if(!good) System.out.println("NO");
    else
    {
        System.out.println("YES");
        for(int i = 0; i<n; i++)
        {
            for(int x : res[i])
                System.out.print(x+" ");
                    System.out.println();
        }
    }
}
}