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 frq[5001];
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> v(n);
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    v[i] = {x, i};
    frq[x]++;
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < 5001; ++i)
    if (frq[i] > k) {
      puts("NO");
      return 0;
    }
  vector<int> ans(n);
  ans[v[0].second] = 1;
  for (int i = 1; i < n; ++i) ans[v[i].second] = (i + 1) % k;
  puts("YES");
  for (auto x : ans)
    if (x == 0)
      cout << k << " ";
    else
      cout << x << " ";
  return 0;
}
