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 PaintingPebbles_B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int arr[] = new int[n];
		int temp[] = new int[n];
		for(int i = 0; i<n; i++) {
			arr[i] = sc.nextInt();
			temp[i] = arr[i];
		}
		
		Arrays.sort(temp);
		int ans[][] = new int[n][temp[n-1]];
		if(k<temp[n-1]-temp[0]) {
			System.out.println("NO");
		}
		else {
			System.out.println("YES");
			int x = 1;
			int aa = 0;
			for(int i = 0; i<temp[n-1]; i++) {
				for(int j = 0; j<n; j++) {
					ans[j][i] = x;
					
				}
				if(i>=temp[0]) {
					x++;
				}
			}
			for(int j = 0; j<n; j++) {
				for(int i = 0; i<temp[n-1] && i<arr[j]; i++) {
					System.out.print(ans[j][i]+" ");
				}
				System.out.println();
			}
			
		}
		

	}

}