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 n, k;
int a[6666], ans[6666];
set<int> s[6666];
void cloro(int id, int clo) {
  while (s[a[id]].count(clo)) {
    clo++;
    if (clo == k + 1) {
      clo = 1;
    }
  }
  ans[id] = clo;
  s[a[id]].insert(clo);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int now = 1;
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    if (s[a[i]].size() == k) {
      cout << "NO" << endl;
      exit(0);
    }
    cloro(i, now);
    now++;
    if (now == k + 1) {
      now = 1;
      flag = true;
    }
  }
  if (!flag && now < k) {
    cout << "NO" << endl;
    exit(0);
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
}
