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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int cases, n, k;
  cases = 1;
  while (cases--) {
    cin >> n >> k;
    vector<pair<int, int>> v(n);
    for (long long int i = 0; i < n; i++) cin >> v[i].first, v[i].second = i;
    sort(v.begin(), v.end());
    bool flag = true;
    for (long long int i = 0; i < n - 1; i++) {
      int c = 0;
      while (i + 1 < n && v[i].first == v[i + 1].first) c++, i++;
      if (c + 1 > k) {
        flag = false;
        break;
      }
    }
    if (flag) {
      cout << "YES"
           << "\n";
      for (long long int i = 0; i < n; i++) {
        v[i].first = v[i].second;
        v[i].second = ((i + 1) % k) + 1;
      }
      sort(v.begin(), v.end());
      for (long long int i = 0; i < n; i++) cout << v[i].second << " ";
    } else
      cout << "NO"
           << "\n";
  }
  return 0;
}
