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 Div2;

import java.util.Scanner;

public class CF290 {
	public static void main(String[] args){
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		
		int[] piles = new int[n];
		for(int i=0; i<piles.length; i++){
			piles[i] = input.nextInt();
		}
		
		int maxi = Integer.MIN_VALUE;
		int mini = Integer.MAX_VALUE;
		
		for(int i=0; i<piles.length; i++){
			if(piles[i] < mini){
				mini = piles[i];
			}
			if(piles[i] > maxi){
				maxi = piles[i];
			}
		}
		
		if(maxi - mini > k){
			System.out.println("NO");
		}
		else {
			int[] shares = new int[k];
			int quo = mini / k;
			int res = mini % k;
			for(int i=0; i<k-res; i++)
				shares[i] = quo;
			for(int i=k-res; i<k; i++)
				shares[i] = quo+1;
			System.out.println("YES");
			for(int i=0; i<n; i++){
				int tmp = piles[i];
				int gap = tmp - mini;
				for(int j=0; j<gap; j++){
					for(int t=0; t<=shares[j]; t++){
						System.out.print((j+1) + " ");
					}
				}
				for(int j=gap; j<k; j++){
					for(int t=0; t<shares[j]; t++){
						System.out.print((j+1) + " ");
					}
				}
				System.out.println();
			}
		}
	}
}