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())
li=list(map(int,input().split()))
a=[0 for _ in range(5001)]
d={}
for i in li:
	a[i]+=1

j=0
flag1=0
flag2=0
o=1
while j<5001:
	if a[j]>k:
		flag1=1
		break

	elif a[j]>=1:
		d[j]=[]
		for i in range(a[j]):
			if o>k:
				o=1
				flag2=1
			d[j].append(o)
			o+=1
		if o>k:
			o=1
			flag2=1
			
	j+=1

if flag1==0 and flag2==1:
	print("YES")
	for i in li:
		print(d[i][0],end=" ")
		d[i].pop(0)
	print()

else:
	print("NO")

			

