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;
  cin >> n >> k;
  int i, b[5001] = {0};
  pair<int, int> a[n];
  for (i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
    b[a[i].first]++;
  }
  pair<int, int> v[n];
  sort(a, a + n);
  int cl = 0;
  for (i = 0; i < n; i++) {
    if (b[a[i].first] > k) {
      cout << "NO" << '\n';
      return 0;
    }
    cl++;
    cl = cl % k;
    if (cl == 0) {
      v[i].first = a[i].second;
      v[i].second = k;
      cl = 0;
    } else {
      v[i].first = a[i].second;
      v[i].second = cl;
    }
  }
  sort(v, v + n);
  cout << "YES" << '\n';
  for (i = 0; i < n; i++) cout << v[i].second << " ";
}
