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;
using LL = long long;
using uLL = unsigned long long;
int close_sycn = (ios::sync_with_stdio(false), cin.tie(0), 0);
constexpr int N = 5005;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n), cnt(N);
  for (auto &x : a) {
    cin >> x;
    cnt[x]++;
  }
  if (*max_element(cnt.begin(), cnt.end()) > k) {
    cout << "NO\n";
    exit(0);
  }
  cout << "YES\n";
  vector<set<int>> s(N);
  for (int i = 0; i < k; i++) {
    cout << i + 1 << ' ';
    s[a[i]].insert(i + 1);
  }
  auto fun = [&](set<int> &s) {
    int b = 1;
    for (auto &x : s) {
      if (b != x) break;
      b++;
    }
    return b;
  };
  for (int i = k; i < n; i++) {
    int t = fun(s[a[i]]);
    cout << t << ' ';
    s[a[i]].insert(t);
  }
  cout << '\n';
  return 0;
}
