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 Main{
 
 
  private final static PrintWriter os = new PrintWriter(System.out);
 
 
  public static void main(String[] args) throws IOException{
    int n = R.B(), q = R.B(), i, j = 0; Boolean f = true, f1 = false;
    ArrayList<ArrayList<Integer>> c = new ArrayList<ArrayList<Integer>>();
    for(i = 0; i<q; i++){c.add(new ArrayList<Integer>());}
    int[] a = new int[n]; int[] an = new int[n];
 
 
    for(i = 0; i<n; i++){
      a[i] = R.B();
      if(f){
        if(!f1){
          j = 0;
          while(j<q){if(c.get(j).isEmpty()){
           c.get(j).add(a[i]); an[i] = j+1; break;
          }else{j++;}}
          if(j == q){f1 = true;
            j = 0;
          while(j<q){if(c.get(j).contains(a[i])){j++;}else{
            c.get(j).add(a[i]); an[i] = j+1; break;
          }}
          if (j == q){f = false;}
          }
        }else{
          j = 0;
          while(j<q){if(c.get(j).contains(a[i])){j++;}else{
            c.get(j).add(a[i]); an[i] = j+1; break;
          }}
          if (j == q){f = false;}
        }
      }
    }


    if(f){os.write("YES\n");
    for(i=0; i<n; i++){os.write(Integer.toString(an[i]) + " ");}
    }else{os.write("NO\n");}

 
    os.flush();
  }
}
 
 
final class R{
  public static BufferedReader a = new BufferedReader(new InputStreamReader(System.in));
  private static StringTokenizer b;
  public static String A() throws IOException{while (b == null || !b.hasMoreTokens()) try {b = new StringTokenizer(a.readLine());} catch (Exception e){e.printStackTrace();}
    return b.nextToken();}
  public static int B() throws IOException{return Integer.parseInt(A());}
  public static String C() throws IOException{return a.readLine();}
}