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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int k = scanner.nextInt();
		int[]a = new int[n];
		Map<Integer, Integer> map = new HashMap<Integer, Integer>();
		for (int i = 0; i < n; i++) {
			a[i] = scanner.nextInt();
			map.put(i, a[i]);
		}
		int num = arraycount(a);
		if(k<num){
			System.out.println("NO");
			return ;
		}
		System.out.println("YES");
		List<Map.Entry<Integer, Integer>> list = new ArrayList<Map.Entry<Integer,Integer>>(map.entrySet());
		Collections.sort(list,new Comparator<Map.Entry<Integer, Integer>>() {

			@Override
			public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) {
				return o1.getValue().compareTo(o2.getValue());
			}
		});
		int temp=1;
		for(Map.Entry<Integer, Integer> maping:list){
			map.replace(maping.getKey(), maping.getValue(), temp);
			if (temp==k) {
				temp = 0;
			}
			temp++;
//			System.out.println(maping.getKey() + ":" + maping.getValue());
		}
		Collections.sort(list,new Comparator<Map.Entry<Integer, Integer>>() {

			@Override
			public int compare(Entry<Integer, Integer> o1, Entry<Integer, Integer> o2) {
				return o1.getKey().compareTo(o2.getKey());
			}
		});
		for(Map.Entry<Integer, Integer> maping:list){
			System.out.print(maping.getValue()+" ");
		}
		
		

	}
	
	public static int arraycount(int [] arr) {
		//获取最大值和最小
		int min = arr[0];
		int max = arr[0];
		for(int i = 0; i < arr.length; i ++) {
			if(min > arr[i]) { min = arr[i];}
			if(max < arr[i]) { max = arr[i];}
		}
	
		// 定义存放数字个数的数组
		int[] counts = new int[max -min + 1];
		int offset = min - 0;// 
		//统计每个数次出现的次数
		for(int i = 0;i < arr.length; i ++) {
			counts[arr[i] - offset] ++;//实现原数组中数字存入计数数组的位置
		}                //数字所对应计算数组的脚标= 原数字 - 偏移量
		
		//遍历计数数组，将每个元素打印出来     偏移量 = offset
		//计算数组中所对应的数是： 偏移量 + 脚标
		int max_count = 0;
		for(int i = 0; i < counts.length;i ++) {  //遍历计数数组
			if( counts[i] != 0) { 
				if(counts[i]>=max_count)
					max_count=counts[i];
//				System.out.println( (offset + i)+ "出现的次数是：" + counts[i]); 
			}
		}
		return max_count;
	}
}

 	  	  	 		  			  	  	   			 		