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 {
    static int maxfun(int a[]){
        int max=Integer.MIN_VALUE;
        for(int i=0;i<a.length;i++){
            if(a[i]>max)
                max=a[i];
        }
        return max;
    }
    static int minfun(int a[]){
        int min=Integer.MAX_VALUE;
        for(int i=0;i<a.length;i++){
            if(a[i]<min)
                min=a[i];
        }
        return min;
    }
    public static void main(String args[]){
        Scanner sc= new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        int a[]=new int[n];
        for(int i=0;i<n;i++)
            a[i]=sc.nextInt();
        int kd=maxfun(a)-minfun(a);
        if(k<kd)
            System.out.println("NO");
        else{
            System.out.println("YES");
            for(int i=0;i<n;i++){
                int l=1;
                for(int j=0;j<Math.min(a[i], minfun(a)+1);j++)
                    System.out.print(l+" ");
                l++;
                if(a[i]>minfun(a)+1){
                    for(int p=0;p<a[i]-(minfun(a)+1);p++,l++)
                        System.out.print(l+" ");
                }
                System.out.println();
            }
        }
    }
}