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>
using namespace std;
int main() {
  int n, k, cnt = 1;
  int num[5005], res[5005], vis[5005];
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    if (i == 1) {
      res[i] = cnt;
    } else {
      int j, ans = 0;
      memset(vis, 0, sizeof(vis));
      for (j = 1; j < i; j++) {
        if (num[i] == num[j]) {
          if (!vis[res[j]]) {
            ans++;
          }
          vis[res[j]]++;
        }
      }
      if (ans) {
        if (ans == cnt) {
          res[i] = ++cnt;
        } else {
          for (j = 1; j <= cnt; j++) {
            if (!vis[j]) {
              res[i] = j;
              break;
            }
          }
        }
      } else {
        res[i] = res[1];
      }
    }
  }
  if (cnt > k) {
    printf("NO\n");
  } else {
    printf("YES\n");
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      if (cnt < k) {
        if (vis[res[i]]) {
          cnt++;
          res[i] = cnt;
        }
      }
      printf("%d", res[i]);
      vis[res[i]] = 1;
      if (i == n) {
        printf("\n");
      } else {
        printf(" ");
      }
    }
  }
  return 0;
}
