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;
bool cmpDouble(const double& a, const double& b) {
  return std::fabs(a - b) < std::numeric_limits<double>::epsilon();
}
int main() {
  int n, k;
  cin >> n >> k;
  int arr[5001];
  int ans[5001];
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    ans[i] = 0;
  }
  map<int, set<int>> seen;
  for (int i = 0; i < k; ++i) {
    ans[i] = i + 1;
    seen[arr[i]].insert(i + 1);
  }
  for (int i = k; i < n; ++i) {
    if (seen[arr[i]].size() >= k) {
      cout << "NO" << endl;
      return 0;
    }
    for (int j = 1; j <= k; ++j) {
      if (seen[arr[i]].find(j) == seen[arr[i]].end()) {
        ans[i] = j;
        seen[arr[i]].insert(j);
        break;
      }
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
