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() {
  int n, k, count(0);
  cin >> n >> k;
  vector<int> v(n), ans(n);
  vector<int> temp(5001);
  for (int i = 0; i < n; ++i) cin >> v[i];
  for (int i = 0; i < n; ++i) {
    ++temp[v[i]];
    if (temp[v[i]] > k) {
      cout << "NO";
      return 0;
    }
  }
  for (int i = 1; i <= k; ++i) {
    vector<int> used(5001);
    bool ext(0);
    for (int j = 0; j < n; ++j) {
      if (!used[v[j]] && !ans[j]) {
        ans[j] = i;
        used[v[j]] = 1;
        ++count;
        if (n - count == k - i) {
          count = i + 1;
          ext = 1;
          break;
        }
      }
    }
    if (ext) break;
  }
  cout << "YES\n";
  for (int i = 0; i < n; ++i) {
    if (!ans[i]) {
      ans[i] = count;
      ++count;
    }
    cout << ans[i] << ' ';
  }
  return 0;
}
