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())
a = list(map(int,input().split()))

b = []
index = {}

for i in range(n):
	b.append(0)
	index[a[i]] = []

for i in range(n):
	index[a[i]].append(i)

ans = 0
check = []
for i in range(n):
	b.append(0)
	check.append(0)

count = 1
for i in range(n):
	# print(i,check[i])
	if(check[i] == 0):
		start = count
		# print(start)
		first = 0
		# print(a[i],index[a[i]])
		for c in index[a[i]]:
			check[c] = 1
			# print(count)
			if first==1 and count == start:
				print("NO")
				ans = 1
				break
			first = 1

			b[c] = count
			count += 1
			count = count%k
			if count == 0:
				count = k
			# print(c,count)


		if ans == 1:
			break

if ans!=1:
	print("YES")
	for c in range(n):
		print(b[c],end= " ")

				
