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 B289 {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		int a[] = new int [n];
		for (int i=0;i<n;i++) {
			a[i] = in.nextInt();
		}
		int b[][] = new int[n][k+1];
		for (int i=1;i<=k;i++) {
			int min = 0;
			if (i==1) {
				min = 200;
				for (int j=0;j<n;j++) {
					min = Math.min(a[j], min);
					
				}
				if (min == 200) break;
			}

			
			min++;
			for (int j=0;j<n;j++) 
			if (a[j]!=0){
				int toTake = Math.min(a[j], min);
				a[j]-=toTake;
				b[j][i] = toTake;
			}
		}
		for (int i=0;i<n;i++) 
			if (a[i]!=0) {
				System.out.println("NO");
				return;
			}
		System.out.println("YES");
		for (int i=0;i<n;i++) {
			for (int j=1;j<=k;j++) {
				for (int o=0;o<b[i][j];o++) {
					System.out.print(j+" ");
				}
			}
			System.out.println();
		}
	}
}