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 fun() {}
const long long mod = 1e9 + 7;
signed main() {
  fun();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1;
  while (tests--) {
    long long n;
    cin >> n;
    long long k;
    cin >> k;
    long long a[n];
    set<long long> s[5007];
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long i = 0; i < n; i++) {
      long long f = 0;
      for (long long j = 0; j < k; j++) {
        if (s[j].find(a[i]) == s[j].end()) {
          s[j].insert(a[i]);
          f = 1;
          break;
        }
      }
      if (f == 0) return cout << "NO\n", 0;
    }
    for (long long i = 0; i < k; i++) {
      if (s[i].size() == 0) {
        long long x = -1, f = 0;
        for (long long j = 0; j < k; j++) {
          if (s[j].size() > 1) {
            x = *s[j].begin();
            s[j].erase(s[j].begin());
            f = 1;
            break;
          }
        }
        if (f == 0) return cout << "NO\n", 0;
        s[i].insert(x);
      }
    }
    cout << "YES\n";
    for (long long i = 0; i < n; i++) {
      long long x = a[i];
      for (long long j = 0; j < k; j++) {
        if (s[j].find(x) != s[j].end()) {
          cout << j + 1 << ' ';
          s[j].erase(x);
          break;
        }
      }
    }
  }
  return 0;
}
