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.
from sys import stdin, stdout
from math import *
from heapq import *
from collections import *

def main():
    n,k=[int(x) for x in stdin.readline().split()]
    a=[int(x) for x in stdin.readline().split()]
    c=[]
    d=[]
    for i in range(5002):
        c.append([])
        d.append(i)
    for i in range(n):
        c[a[i]].append(i)
        if (len(c[a[i]])>k):
            stdout.write("NO")
            return 0
    d.sort(key=lambda x: len(c[x]),reverse=True)
    i=0
    res=[""]*n
    for idc in d:
        for ida in c[idc]:
            res[ida]=str(i+1)
            i=(i+1) % k
    stdout.write("YES\n")
    stdout.write(" ".join(res))
    return 0

if __name__ == "__main__":
    main()