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
import java.io.BufferedReader;
import java.io.*;
import java.util.*;

public class Test {
    
    
    public static void main(String[] args) throws Exception {
        
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String s[] = in.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int k = Integer.parseInt(s[1]);
        s = in.readLine().split(" ");
        int a[] = new int[n];
        for(int i = 0 ; i < n; i++){
            a[i] = Integer.parseInt(s[i]);
        }
        int ans[][] = new int[n][k];
        int max[] = new int[k];
        int div = a[0]/k;
        int re = a[0]%k;
        for(int i = 0; i < k; i++)
        {    
            ans[0][i] = div;
            if(re > 0){
                ans[0][i]++;
                re--;
            }
            max[i] = ans[0][i];
        }
        for(int i = 1; i < n; i++){
            div = a[i]/k;
            re = a[i]%k;
            for(int j = 0; j < k; j++){
                ans[i][j] = div;
                if(re > 0){
                    int dif = Math.abs(max[j] - (ans[i][j]+1));
                    if(dif <= 1){
                        ans[i][j]++;
                        re--;
                    }
                }
                int dif =  Math.abs(max[j] - (ans[i][j]));
                if(dif > 1)
                {
                    System.out.println("NO");
                    return;
                }
                max[j] = Math.max(max[j],ans[i][j]);
            }
            if(re > 0)
            {
                System.out.println("NO");
                return;
            }
            // System.out.println(Arrays.toString(ans[i]));
        }
        StringBuilder sb = new StringBuilder();
        sb.append("YES\n");
        for(int i = 0 ; i < n; i++){
            for(int j = 0 ; j < k; j++){
                 int l = ans[i][j];
                 while(l-- > 0){
                     sb.append((j+1)+" ");
                 }
            }
            sb.append("\n");
        }
        System.out.print(sb.toString());
    }
}