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;
  scanf("%d %d", &n, &k);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  bool ans = true;
  int c[n];
  for (int i = 0; i < k; i++) {
    c[i] = i;
  }
  for (int i = k; i < n; i++) {
    bool v[k];
    memset(v, false, sizeof(v));
    for (int j = 0; j < i; j++) {
      if (a[j] == a[i]) {
        v[c[j]] = true;
      }
    }
    int tc = -1;
    for (int j = 0; j < k; j++) {
      if (!v[j]) {
        tc = j;
        break;
      }
    }
    if (tc >= 0) {
      c[i] = tc;
    } else {
      ans = false;
      break;
    }
  }
  if (ans) {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      printf("%d%c", c[i] + 1, i == n - 1 ? '\n' : ' ');
    }
  } else {
    printf("NO\n");
  }
  return 0;
}
