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;
const int N = 5001;
vector<int> g[N];
int color[N], n;
size_t k, l = 1;
inline int solution(bool yes) {
  const string ans[] = {"NO", "YES"};
  cout << ans[yes];
  if (yes) {
    cout << '\n' << color[0];
    for (int i = 1; i < n; i++) cout << ' ' << color[i];
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr),
      cin >> n >> k;
  for (int a, i = 0; i < n; i++) cin >> a, g[a].push_back(i);
  for (int i = 1; i < N; i++)
    if (g[i].size() <= k)
      for (auto j : g[i]) color[j] = l, l %= k, l++;
    else
      return solution(false);
  return solution(true);
}
