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.Arrays;
import java.util.Scanner;


public class B509 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        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 min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            min = Math.min(min, a[i]);
            max = Math.max(max, a[i]);
        }
        
        int[] bMin = new int[min + 1];
        int[] bMax = new int[max + 1];

        int[] cMin = new int[k];
        int[] cMax = new int[k];
        Arrays.fill(cMin, 0);
        Arrays.fill(cMax, 0);
        
        for (int j = 1; j <= min; j++) {
            int c = j % k;
            if (c == 0)
                bMin[j] = k;
            else
                bMin[j] = c;
            cMin[bMin[j] - 1]++;
        }
        
        for (int j = 1; j <= max; j++) {
            int c = j % k;
            if (c == 0)
                bMax[j] = k;
            else
                bMax[j] = c;
            cMax[bMax[j] - 1]++;
        }
        
        for (int c = 0; c < k; c++)
            if (cMax[c] - cMin[c] > 1) {
                System.out.println("NO");
                return;
            }
        
        System.out.println("YES");
        for (int i = 0; i < n; i++) {
            for (int j = 1; j <= a[i]; j++) {
                int c = j % k;
                if (c == 0)
                    System.out.print(k + " ");
                else
                    System.out.print(c + " ");
            }
                
            System.out.println();
        }
    }

}