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:
//package Round289;

import java.util.Scanner;
public class SolutionB {
  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int[] pepples = new int[n];
    for (int i = 0; i < pepples.length; i++)
       pepples[i] = scan.nextInt();
    int max = Integer.MIN_VALUE, min = Integer.MAX_VALUE;
    for (int i = 0; i < pepples.length; i++) {
      if (pepples[i] > max) max = pepples[i];
      if (pepples[i] < min) min = pepples[i];
    }
    if (max - min > k) 
      System.out.println("NO");
    else {
      System.out.println("YES");
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < min; j++) {
          System.out.print("1 ");
        }
        for (int j = min, m = 1; j < pepples[i]; j++, m++) {
          System.out.print(m + " ");
        }
        System.out.print("\n");
      }
    }
    
    scan.close();
  }
}