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 int;
const int MOD = 1e9 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t-- > 0) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    map<int, vector<int>> mp;
    bool poss = true;
    for (int i = 0; i < int(n); i++) {
      cin >> a[i];
      mp[a[i]].push_back(i);
      if (poss && mp[a[i]].size() > k) poss = false;
    }
    if (!poss) {
      cout << "NO\n";
    } else {
      vector<int> colors(k);
      for (int i = 0; i < int(k); i++) colors[i] = i + 1;
      vector<int> ans(n);
      int color = 0;
      for (const auto& i : mp) {
        vector<int> tofill = i.second;
        for (auto j : tofill) {
          ans[j] = colors[color];
          color++;
          if (color == k) color = 0;
        }
      }
      cout << "YES\n";
      for (auto i : ans) {
        cout << i << " ";
      }
      cout << '\n';
    }
  }
}
