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() {
  long long int n, k, cnt = 0, x = 0, start = 0, y = 0;
  cin >> n >> k;
  pair<long long int, long long int> p[n + 1];
  long long int a[n + 1], i;
  p[0].first = 0;
  p[0].second = 0;
  for (i = 1; i <= n; i++) {
    cin >> p[i].first;
    p[i].second = i;
  }
  sort(p, p + n + 1);
  for (i = 1; i <= n; i++) {
    if (p[i].first == x) {
      cnt++;
      y = ((start + cnt) % k);
      if (y == 0) y = k;
      a[p[i].second] = y;
      if (cnt == k) {
        cout << "NO";
        return 0;
      }
    } else {
      x = p[i].first;
      start = y + 1;
      if (start > k) start = 1;
      cnt = 0;
      y = ((start + cnt) % k);
      if (y == 0) y = k;
      a[p[i].second] = y;
    }
  }
  cout << "YES" << endl;
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}
