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;
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
const long long int mod = 1000000007;
const long long int inf = (long long int)1e15;
long long int power(long long int a, long long int b, long long int m = mod) {
  if (b == 0) return 1;
  if (b == 1) return (a % m);
  long long int x = power(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2) x = (x * a) % m;
  return x;
}
long long int max(long long int a, long long int b) { return (a > b ? a : b); }
long long int min(long long int a, long long int b) { return (a < b ? a : b); }
const int N = 5111;
int n, k;
int a[N];
vector<int> v[N];
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v[a[i]].push_back(i);
  }
  for (int i = 1; i < N; i++) {
    if (v[i].size() > k || k > n) {
      cout << "NO"
           << "\n";
      ;
      return;
    }
  }
  int c[N] = {0};
  int col = 0;
  for (int i = 1; i < N; i++) {
    if (v[i].size() > 0) {
      for (int j = 0; j < v[i].size(); j++) {
        c[v[i][j]] = col++;
      }
    }
  }
  if (col < k) {
    cout << "NO"
         << "\n";
    ;
    return;
  }
  cout << "YES"
       << "\n";
  ;
  for (int i = 1; i <= n; i++) {
    cout << c[i] % k + 1 << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1, num = 1;
  while (t--) {
    solve();
  }
  return 0;
}
