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;
  cin >> n >> k;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  map<int, int> m;
  for (int i = 0; i < n; i++) {
    if (m.find(v[i]) == m.end()) {
      m.insert({v[i], 1});
    } else {
      int l = m[v[i]];
      m[v[i]] = l + 1;
    }
  }
  int maxi = INT_MIN;
  for (map<int, int>::iterator it = m.begin(); it != m.end(); ++it)
    maxi = max(maxi, it->second);
  if (k > n || k < maxi) {
    cout << "NO" << endl;
  } else {
    vector<int> ans(n, -1);
    int b = 1;
    for (int i = 0; i < n; i++) {
      if (ans[i] == -1) {
        ans[i] = b;
        for (int j = i + 1; j < n; j++) {
          if (v[j] == v[i]) {
            b = (b + 1) % (k + 1);
            if (b == 0) b = 1;
            ans[j] = b;
          }
        }
        b = (b + 1) % (k + 1);
        if (b == 0) b = 1;
      }
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
