You are given an array a consisting of n integer numbers.

You have to color this array in k colors in such a way that: 

  * Each element of the array should be colored in some color; 
  * For each i from 1 to k there should be at least one element colored in the i-th color in the array; 
  * For each i from 1 to k all elements colored in the i-th color should be distinct. 



Obviously, such coloring might be impossible. In this case, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions above. If there are multiple answers, you can print any.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the length of the array a and the number of colors, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5000) — elements of the array a.

Output

If there is no answer, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions described in the problem statement. If there are multiple answers, you can print any.

Examples

Input


4 2
1 2 2 3


Output


YES
1 1 2 2


Input


5 2
3 2 1 2 3


Output


YES
2 1 1 2 1


Input


5 2
2 1 1 2 1


Output


NO

Note

In the first example the answer 2~ 1~ 2~ 1 is also acceptable.

In the second example the answer 1~ 1~ 1~ 2~ 2 is also acceptable.

There exist other acceptable answers for both examples.
import java.util.Scanner;


public class Main {



	public static void sortIndex(Pair[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (a[i].index > a[j].index) {
					Pair t = a[i];
					a[i] = a[j];
					a[j] = t;
				}
			}
		}
	}

	public static void sortValue(Pair[] a) {

		int n = a.length;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (a[i].value > a[j].value) {
					Pair t = a[i];
					a[i] = a[j];
					a[j] = t;
				}
			}
		}
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		Pair[] a = new Pair[n];

		for (int i = 0; i < n; i++) {
			Pair t = new Pair(i,in.nextInt(), -1 );
			a[i] = t;
		}
		sortValue(a);
		int b = 1;
		int max = Integer.MIN_VALUE;
		for (int i = 0; i < n; i++) {
			a[i].colour = i % k;
			if (i > 0) {
				if (a[i].value == a[i - 1].value)
					b++;
				else {
					max = Math.max(max, b);
					b = 1;
				}
			}
			if (b > k) {
				System.out.println("NO");
				System.exit(0);
				break;
			}
		}
		sortIndex(a);
		System.out.println("YES");
		for (int i = 0; i < n; i++) {
			int t = a[i].colour + 1;
			System.out.print(t + " ");
		}
	}
	public static  class Pair {
		public int index;
		public int value;
		public int colour;
		
		public Pair(int a, int b, int c) {
			index = a;
			value = b;
			colour = c;
		}
	}
	}
	


