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>
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
static const long double epsilon = 1e-15;
static const long double pi = acos((long double)-1);
using namespace std;
inline void init_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int arr[5001];
int narr[5001];
int occ[5001];
int m[5001];
int main() {
  int n, k;
  cin >> n >> k;
  int mx = 0;
  for (int(i) = (0); (i) < (n); i++) {
    cin >> arr[i];
    m[arr[i]]++;
    narr[i] = m[arr[i]];
    occ[narr[i]]++;
    mx = max(narr[i], mx);
  }
  if (mx > k)
    cout << "NO" << '\n';
  else {
    cout << "YES" << '\n';
    if (mx == k) {
      for (int(i) = (0); (i) < (n); i++) cout << narr[i] << " ";
      return 0;
    }
    mx++;
    for (int(i) = (1); (i) < (k); i++) {
      if (occ[i] > 1) {
        int kk = occ[i];
        for (int(j) = (1); (j) < (n); j++) {
          if (kk == 1) break;
          if (narr[j] == i) {
            narr[j] = mx;
            mx++;
            kk--;
            if (mx > k) {
              for (int(i) = (0); (i) < (n); i++) cout << narr[i] << " ";
              return 0;
            }
          }
        }
      }
    }
    for (int(i) = (0); (i) < (n); i++) cout << narr[i] << " ";
  }
  return 0;
}
