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_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, k;
    cin >> n >> k;
    vector<pair<long long int, long long int> > v(n);
    long long temp;
    for (long long int i = 0; i < n; i++) {
      cin >> v[i].first;
      v[i].second = i;
    }
    sort((v).begin(), (v).end());
    long long cnt = 1;
    long long mcnt = 0;
    for (long long int i = 1; i < n; i++) {
      if (v[i].first != v[i - 1].first) {
        if (cnt > mcnt) mcnt = cnt;
        cnt = 1;
      } else
        cnt++;
    }
    if (cnt > mcnt) mcnt = cnt;
    if (mcnt > k)
      cout << "NO" << '\n';
    else {
      cout << "YES" << '\n';
      vector<long long int> x(n);
      long long i = 1;
      for (long long int j = 0; j < n; j++) {
        x[v[j].second] = i;
        i++;
        if (i == k + 1) i = 1;
      }
      for (long long int i = 0; i < n - 1; i++) cout << x[i] << " ";
      cout << x[n - 1];
    }
  }
  return 0;
}
