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;
deque<long long int> d;
vector<long long int> v1;
map<long long int, long long int> m;
vector<pair<long long int, long long int> > v;
long long int i, j, n, p, q, a, k, t;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  long long int b[n];
  for (i = 0; i < n; i++) {
    cin >> a;
    v.push_back(make_pair(a, i));
    m[a]++;
    p = max(p, m[a]);
  }
  sort(v.begin(), v.end());
  b[v[0].second] = 1;
  if (p > k)
    cout << "NO";
  else {
    cout << "YES" << endl;
    for (i = 1; i < n; i++) {
      t = v[i].second;
      q = v[i - 1].second;
      if (b[q] == k) {
        b[t] = 1;
      } else
        b[t] = b[q] + 1;
    }
    for (i = 0; i < n; i++) cout << b[i] << ' ';
  }
}
