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>
#pragma GCC optimize("O2")
using namespace std;
int power(int x, unsigned int y) {
  int res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<int> v(5001, 0);
  vector<int> vv(n);
  vector<int> final;
  multimap<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> vv[i];
    final.push_back(vv[i]);
    v[vv[i]]++;
  }
  int max = *max_element(v.begin(), v.end());
  if (max > k) {
    cout << "NO";
    return 0;
  } else {
    cout << "YES\n";
    sort(vv.begin(), vv.end());
    int t = 0;
    for (int i = 0; i < n; i++) {
      t++;
      t = t % (k + 1);
      if (t == 0) {
        t = 1;
      }
      mp.insert({vv[i], t});
    }
    for (int i = 0; i < n; i++) {
      auto itr = mp.find(final[i]);
      cout << itr->second << " ";
      mp.erase(itr);
    }
  }
  return 0;
}
