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 a[5005];
int vis[5005];
int b[5005];
int num[5005];
int main() {
  int n, k;
  int f = 1;
  int MAX = 0;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; i++) {
    b[i] = i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    vis[a[i]]++;
    if (vis[a[i]] > MAX) {
      MAX = vis[a[i]];
    }
    if (vis[a[i]] > k) {
      f = 0;
    } else {
      a[i] = b[vis[a[i]]];
      num[a[i]]++;
    }
  }
  for (int i = n; i >= 1; i--) {
    if (num[a[i]] > 1 && MAX < k) {
      num[a[i]]--;
      a[i] = b[++MAX];
      num[a[i]]++;
    }
  }
  if (f == 0) {
    printf("NO\n");
  } else {
    printf("YES\n");
    for (int i = 1; i <= n; i++) {
      if (i == n) {
        printf("%d\n", a[i]);
      } else {
        printf("%d ", a[i]);
      }
    }
  }
  return 0;
}
