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.
#include <bits/stdc++.h>
int main() {
  int n, k, a[5005], rec[5005] = {0}, ans[5005] = {0};
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    rec[a[i]]++;
  }
  int num = 0, id = 0;
  for (int i = 0; i < 5005; i++)
    if (num < rec[i]) {
      num = rec[i];
      id = i;
    }
  if (num > k)
    printf("NO");
  else {
    printf("YES\n");
    int s = 0;
    for (int i = 0; i < n; i++)
      if (a[i] != id) {
        if (s >= k - num) break;
        ans[i] = s + num + 1;
        s++;
      }
    for (int i = 0; i < n; i++)
      if (ans[i] == 0) {
        ans[i] = rec[a[i]];
        rec[a[i]]--;
      }
    for (int i = 0; i < n; i++) printf("%d ", ans[i]);
  }
}
