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 main() {
  int n, k, x, i, cnt, flag = 0, arr[6000], temp, color = 0;
  pair<int, int> p;
  multimap<int, pair<int, int> > m;
  map<int, int> mm;
  cin >> n >> k;
  if (k > n) flag = 1;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    mm[arr[i]] += 1;
    p.first = 0;
    p.second = i;
    m.insert(multimap<int, pair<int, int> >::value_type(arr[i], p));
  }
  for (auto it = mm.begin(); it != mm.end(); it++) {
    if ((*it).second > k) {
      flag = 1;
      break;
    }
  }
  if (flag) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
    color = 0;
    for (auto it = m.begin(); it != m.end(); it++) {
      (*it).second.first = color % k + 1;
      color++;
    }
    for (auto it = m.begin(); it != m.end(); it++) {
      arr[(*it).second.second] = (*it).second.first;
    }
    cout << arr[0];
    for (i = 1; i < n; i++) cout << " " << arr[i];
  }
  return 0;
}
