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;
import java.util.*;

public class Programa {
	
	public static boolean isPalindrome(String a){
		for(int i = 0 , j = a.length() - 1; i < a.length(); i++, j--){
			if(a.charAt(i) == a.charAt(j)) continue;
			else return false;
		}
		return true;
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int k = scan.nextInt();
		int [][] mat = new int [n][k];
		for(int i = 0; i < n; i++){
			int p = scan.nextInt();
			for(int j = 0; j < p; j++){
				mat[i][j % k] += 1;
			}
		}
		

		for(int i = 0; i < n; i++){
			
			for(int j = 0; j < k; j++){
				for(int r = i + 1; r < n; r++){
					if(Math.abs(mat[i][j] - mat[r][j]) <= 1) continue;
					else {System.out.println("NO"); System.exit(0);}
				}
			}
		}
		System.out.println("YES");
		for(int i = 0; i < n; i++){
			for(int j= 0; j < k; j++){
				for(int r = 0; r < mat[i][j]; r++)
					System.out.print((j+1) + " ");
			}
			System.out.println();
		}
		
	}
}