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(false);
  cin.tie(NULL);
  long long n, d, k = 1;
  cin >> n >> d;
  vector<long long> v(n), r(n);
  map<long long, long long> mapa;
  vector<vector<long long> > p(5001);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    mapa[v[i]]++;
    p[v[i]].push_back(i);
  }
  for (auto i : mapa) {
    if (i.second > d) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << "YES\n";
  for (long long i = 1; i <= 5000; i++) {
    for (long long j = 0; j < int(p[i].size()); j++) {
      r[p[i][j]] = k;
      if (k == d)
        k = 1;
      else
        k++;
    }
  }
  for (long long i : r) cout << i << " ";
  cout << "\n";
  return 0;
}
