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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long arr[5001] = {0};
  long long *a = new long long[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    arr[a[i]]++;
  }
  bool f = true;
  for (long long i = 1; i <= 5000; i++) {
    if (arr[i] > k) {
      f = false;
    }
  }
  if (f) {
    cout << "YES"
         << "\n";
    vector<long long> *e = new vector<long long>[5001];
    for (long long i = 0; i < n; i++) {
      e[a[i]].push_back(i);
    }
    vector<long long> *p = new vector<long long>[k];
    long long z = 0;
    for (long long i = 0; i < 5001; i++) {
      long long cnt = 0;
      while (cnt < e[i].size()) {
        p[z].push_back(e[i][cnt]);
        cnt++;
        z = (z + 1) % k;
      }
    }
    long long *ans = new long long[n];
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < p[i].size(); j++) {
        ans[p[i][j]] = i + 1;
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << "\n";
    delete[] ans;
    delete[] p;
    delete[] e;
  } else {
    cout << "NO"
         << "\n";
  }
  delete[] a;
  return 0;
}
