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 int n, k;
  cin >> n >> k;
  long long int y = 1;
  long long int a[n];
  long long int col[n];
  memset(col, 0, sizeof(col));
  ;
  map<long long int, long long int> mp, colmp;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (mp[a[i]] == 0) {
      col[i] = 1;
      colmp[1]++;
    }
    mp[a[i]]++;
  }
  bool pos = true;
  for (long long int i = 0; i < n; i++) {
    bool ok = true;
    long long int y = 1;
    for (long long int j = i + 1; j < n; j++) {
      if (a[j] == a[i] && col[j] == 0) {
        y++;
        if (y <= k) {
          col[j] = y;
          colmp[y]++;
        } else {
          ok = false;
          break;
        }
      }
    }
    if (ok == false) {
      pos = false;
      break;
    }
  }
  if (pos) {
    cout << "YES" << endl;
    long long int c = k;
    for (long long int i = 0; i < n; i++) {
      if (colmp[c] == 0) {
        long long int u = col[i];
        if (colmp[u] >= 2) {
          col[i] = c;
          c--;
          colmp[u]--;
        }
      }
    }
    for (long long int i = 0; i < n; i++) cout << col[i] << " ";
    cout << endl;
  } else
    cout << "NO" << endl;
  return 0;
}
