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, m, a[5001], i, j, k, t, l, b[5001], result[5001];
  cin >> n >> k;
  memset(b, 0, sizeof(b));
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]]++;
  }
  t = 0;
  j = 0;
  for (i = 1; i <= 5000; i++)
    if (b[i] > t) {
      t = b[i];
      m = i;
    }
  if (t > k) {
    cout << "NO";
    return 0;
  } else
    cout << "YES" << endl;
  for (i = 1; i <= n; i++) {
    b[a[i]]--;
    result[i] = k - b[a[i]];
  }
  i = 1;
  if (t < k)
    for (j = 1; j <= n; j++) {
      if (m != a[j]) {
        result[j] = i;
        i++;
      }
      if (i > (k - t)) break;
    }
  for (i = 1; i <= n; i++) cout << result[i] << " ";
}
