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;
const int mod = 1e9 + 7;
const int N = 5000 + 5;
long long fast_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return res % mod;
}
int n, k;
int res[N];
bool vis[N];
int main() {
  cin >> n >> k;
  if (n < k) {
    cout << "NO" << endl;
    return 0;
  }
  int cnt = 0, c = 0;
  vector<pair<int, int> > a(n + 1);
  for (int i = 1; i <= n; i++) {
    int tmp;
    cin >> tmp;
    a[i].first = tmp, a[i].second = i;
  }
  sort(a.begin() + 1, a.begin() + 1 + n);
  for (int i = 1; i <= n; i++) {
    cnt++;
    if (cnt > k) {
      cout << "NO" << endl;
      return 0;
    }
    res[a[i].second] = c;
    c = (c + 1) % k;
    if (i + 1 <= n && a[i + 1].first != a[i].first) {
      cnt = 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 1; i <= n; i++) cout << res[i] + 1 << " ";
  return 0;
}
