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 cnt[5005];
int arr[5005];
int kk[5005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int v;
    scanf("%d", &v);
    arr[i] = v;
    cnt[v]++;
  }
  int high_freq = 1;
  for (int i = 1; i <= 5000; i++) {
    if (cnt[i] > cnt[high_freq]) {
      high_freq = i;
    }
    if (cnt[i] > k) {
      printf("NO");
      return 0;
    }
  }
  printf("YES\n");
  kk[5002] = cnt[high_freq];
  if (cnt[high_freq] != k) {
    int diff = k - cnt[high_freq];
    for (int i = 0; i < n; i++) {
      if (arr[i] != high_freq) {
        arr[i] = 5002;
        diff--;
      }
      if (diff == 0) break;
    }
  }
  for (int i = 0; i < n; i++) {
    int a = arr[i];
    printf("%d ", ++kk[a]);
  }
}
