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;
bool func(pair<int, int> f, pair<int, int> s) { return f.second > s.second; }
int main() {
  int n, c;
  cin >> n >> c;
  int t_c = c;
  int a[n];
  int b[n];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  vector<pair<int, int> > v;
  v.push_back(make_pair(b[0], 1));
  for (int i = 1; i < n; ++i) {
    if (b[i] == b[i - 1])
      v.back().second++;
    else
      v.push_back(make_pair(b[i], 1));
  }
  int s[v.size()];
  sort(v.begin(), v.end(), func);
  if (v[0].second > c) {
    cout << "NO";
    return 0;
  }
  s[0] = v[0].second;
  c -= s[0];
  for (int i = 1; i < v.size(); ++i) {
    if (c == 0)
      s[i] = v[i].second;
    else {
      if (c >= v[i].second) {
        s[i] = s[i - 1] + v[i].second;
        c -= v[i].second;
      } else {
        s[i] = t_c;
        c = 0;
      }
    }
  }
  map<int, int> m;
  for (int i = 0; i < v.size(); ++i) {
    m.insert(make_pair(v[i].first, s[i]));
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) {
    cout << m[a[i]] << ' ';
    m[a[i]]--;
  }
  return 0;
}
