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 JAN18 {

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		int n = scn.nextInt();
		int k = scn.nextInt();
		oidata[] carr = new oidata[n];
		for (int i = 0; i < n; i++) {
			carr[i] = new oidata(i, scn.nextInt());
		}
		Kcoloring(carr.length, k, carr);

	}

	public static class oidata {
		int oi;
		int data;

		public oidata(int oi, int data) {
			this.oi = oi;
			this.data = data;
		}
	}

	public static void Kcoloring(int n, int k, oidata[] arr) {
		oidata[] sorted = mergesort(0, arr.length - 1, arr);
		int[] colors = new int[sorted.length];
		int prev = sorted[0].data, counter = 1;
		colors[sorted[0].oi] = 1;
		int color = 1;
		for (int i = 1; i < sorted.length; i++) {
			color++;
			if (color > k) {
				color = 1;
			}
			if (sorted[i].data == prev) {
				counter++;
				colors[sorted[i].oi] = color;
				if (counter > k) {
					System.out.println("NO");
					return;
				}
			} else {
				prev = sorted[i].data;
				counter = 1;
				colors[sorted[i].oi] = color;
			}

		}
		System.out.println("YES");
		for (int c : colors) {
			System.out.print(c + " ");
		}
	}

	public static oidata[] mergesort(int lo, int hi, oidata[] arr) {
		if (lo > hi) {
			return null;
		}
		if (lo == hi) {
			oidata[] ba = new oidata[1];
			ba[0] = new oidata(lo, arr[lo].data);
			return ba;
		}
		int mid = (lo + hi) / 2;
		oidata[] left = mergesort(lo, mid, arr);
		oidata[] right = mergesort(mid + 1, hi, arr);
		oidata merged[] = mergetwosortedarrays(left, right);
		return merged;
	}

	private static oidata[] mergetwosortedarrays(oidata[] left, oidata[] right) {
		oidata[] sorted = new oidata[left.length + right.length];
		int li = 0, ri = 0, k = 0;
		while (li < left.length && ri < right.length) {
			if (left[li].data > right[ri].data) {
				sorted[k] = new oidata(right[ri].oi, right[ri].data);
				ri++;
			} else {
				sorted[k] = new oidata(left[li].oi, left[li].data);
				li++;
			}
			k++;
		}
		while (li < left.length) {
			sorted[k] = new oidata(left[li].oi, left[li].data);
			li++;
			k++;
		}

		while (ri < right.length) {
			sorted[k] = new oidata(right[ri].oi, right[ri].data);
			ri++;
			k++;
		}

		return sorted;
	}
}
