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;
void fail() {
  cout << "NO";
  exit(0);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  long long n, k, cur = 0;
  map<long long, long long> c;
  map<long long, set<long long>> ex;
  vector<long long> mas, res;
  cin >> n >> k;
  mas.resize(n);
  res.assign(n, -1);
  for (long long i = 0; i < n; i++) {
    cin >> mas[i];
  }
  for (long long i = 0; i < n; i++) {
    if (cur == k) break;
    res[i] = cur++;
    ex[mas[i]].insert(res[i]);
  }
  if (cur != k) fail();
  for (long long i = 0; i < n; i++) {
    if (res[i] != -1) continue;
    while (ex[mas[i]].find(c[mas[i]]) != ex[mas[i]].end()) {
      c[mas[i]]++;
    }
    if (c[mas[i]] == k) fail();
    res[i] = c[mas[i]]++;
  }
  cout << "YES\n";
  for (auto i : res) cout << i + 1 << " ";
  return 0;
}
