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() {
  map<long long, long long> C;
  vector<pair<long long, long long>> B;
  long long n, k, m = 0, v = 1;
  cin >> n >> k;
  long long A[n], D[n];
  for (long long i = 0; i < n; i++) {
    cin >> A[i];
    B.push_back(make_pair(A[i], i));
    C[A[i]]++;
  }
  sort(B.begin(), B.end());
  for (auto it = C.begin(); it != C.end(); it++) {
    m = max(m, it->second);
  }
  if (m > k)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    for (long long i = 0; i < n; i++) {
      if (v % k != 0)
        D[B[i].second] = v % k;
      else
        D[B[i].second] = k;
      v++;
    }
    for (long long i = 0; i < n; i++) cout << D[i] << ' ';
    cout << '\n';
  }
  return 0;
}
