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;
long long int i, j, k;
const int N = 200 * 1000 + 9;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (i = 0; i < n; i++) cin >> v[i];
  if (k > n) {
    cout << "NO" << '\n';
    return;
  }
  map<long long int, long long int> mpp;
  vector<long long int> color;
  long long int mx = 0;
  for (i = 0; i < n; i++) mpp[v[i]]++;
  for (i = 0; i < n; i++) {
    if (mpp[v[i]] > k) {
      cout << "NO" << '\n';
      return;
    }
    color.push_back(mpp[v[i]]);
    mx = max(mx, mpp[v[i]]);
    mpp[v[i]]--;
  }
  map<long long int, long long int> color_used;
  for (i = 0; i < n; i++) {
    if (color_used[color[i]]) {
      mx = mx + 1;
      if (mx > k) break;
      color[i] = mx;
    }
    color_used[color[i]]++;
  }
  cout << "YES" << '\n';
  for (i = 0; i < n; i++) cout << color[i] << " ";
  cout << '\n';
  ;
  return;
}
signed main() {
  solve();
  return 0;
}
