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;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  map<long long, vector<long long>> mp;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].push_back(i);
  }
  long long max_col = 0;
  for (auto u : mp) max_col = max(max_col, (long long)u.second.size());
  if (max_col > k) {
    cout << "NO\n";
    return 0;
  }
  vector<long long> ans(n);
  long long cur_col = 1;
  for (auto u : mp) {
    for (auto v : u.second) {
      ans[v] = cur_col++;
      if (cur_col == k + 1) cur_col = 1;
    }
  }
  cout << "YES\n";
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
