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.
n, k = map(int,input().split())
number=list(map(int,input().split()))
dict={}
for index,value in enumerate(number):
    if value not in dict:
        dict[value]=[index]
    else:
        dict[value].append(index)
L=[]
for alli in dict.values():
    for ii in range(len(alli)):
        L.append(alli[ii])
dict_count={}
for key in number:
    dict_count[key] = dict_count.get(key, 0) + 1
maxnum=max(dict_count.values())
if k>=maxnum:
    print("YES")
    k_cur = k
    result=[0]*n
    for i in range(n):
        if k_cur>1:
            result[L[i]]=k_cur
            k_cur -= 1
        else:
            result[L[i]]=1
            k_cur=k
    for j in range(n):
        if j==n-1:
            print(result[j])
        else:
            print(result[j], end=" ")
else:
    print("NO")
			 			 	   	  	   	  		 		 	 	