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 codeforces.cfr289div2;

import java.util.Scanner;

/**
 * Created by raggzy on 31-Jan-15.
 */
public class B {
    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[][] color = new int[n][];
        int[][] count = new int[n][k];
        // one by one distribution
        for (int i = 0; i < n; i++) {
            color[i] = new int[a[i]];
            int c = 0;
            for (int pile = 0; pile < a[i]; pile++) {
                count[i][c]++;
                color[i][pile] = c + 1;
                c = (c + 1) % k;
            }
        }
        // check if possible
        boolean possible = true;
        for (int c = 0; c < k; c++) {
            int min = Integer.MAX_VALUE;
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < n; i++) {
                min = Math.min(min, count[i][c]);
                max = Math.max(max, count[i][c]);
            }
            if (max - min > 1) {
                possible = false;
                break;
            }
        }

        if (possible) {
            System.out.println("YES");
            for (int i = 0; i < n; i++) {
                StringBuilder sb = new StringBuilder();
                sb.append(color[i][0]);
                for (int pile = 1; pile < a[i]; pile++) {
                    sb.append(' ').append(color[i][pile]);
                }
                System.out.println(sb.toString());
            }
        } else {
            System.out.println("NO");
        }
    }
}