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.io.*;
import java.util.*;
public class div531B
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;

	public static void main(String[] args) throws IOException {
		new div531B().run();
	}
	void run() throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve() throws IOException {
		int n = ni();
		int k = ni();
		int a[] = new int[n+1];
		int cnt[] = new int[5002];
		int max = 0;
		for(int i=1 ; i<=n ; i++) {
			a[i] = ni();
			++cnt[a[i]];
			max = Math.max( max , cnt[a[i]] );
		}
		if( max > k )
			ob.println("NO");
		else {
			boolean good = true;
			int color = 1;
			int res[] = new int[n+1];
			Arrays.fill( res , -1 );
			for(int i=1 ; i<=n ; i++) {
				if( res[i] == -1 ) {
					for(int j=1 ; j<=n ; j++) {
						if( res[j] == -1 && a[j] == a[i] )
							{
								res[j] = color++;
								if( color == k+1 ) {
									color = 1;
									good = true;
								}
							}
					}
				}
			}
			if( good ){
				ob.println("YES");
				for(int i=1; i<=n ; i++)
				ob.print(res[i]+" ");
			} else {
				ob.println("NO");
			}
			
		}
	}
	String ns() throws IOException {
		return nextToken();
	}
	long nl() throws IOException {
		return Long.parseLong(nextToken());
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	double nd() throws IOException {
		return Double.parseDouble(nextToken());
	} 
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	int[] nia(int start,int b) throws IOException {
		int a[]=new int[b];
		for(int i=start;i<b;i++)
			a[i]=ni();
		return a;
	}
	long[] nla(int start,int n) throws IOException {
		long a[]=new long[n];
		for (int i=start; i<n ;i++ ) {
			a[i]=nl();
		}
		return a;
	}
}