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;
long long int const N = 5500;
vector<long long int> vec, adj[6000];
long long int br[6000];
void solve(long long int n, long long int k) {
  long long int z = 1, x, m = k + 1;
  for (long long int i = 1; i <= N; i++) {
    if (adj[i].size() > 1) {
      for (long long int j = 0; j < adj[i].size(); j++) {
        x = adj[i][j];
        z %= m;
        if (z == 0) z = 1;
        br[x] = z;
        z++;
      }
    } else if (adj[i].size() == 1) {
      vec.push_back(adj[i][0]);
    }
  }
  for (long long int i = 0; i < vec.size(); i++) {
    x = vec[i];
    z %= m;
    if (z == 0) z = 1;
    br[x] = z;
    z += 1;
  }
  cout << "YES" << endl;
  for (long long int i = 1; i <= n; i++) {
    cout << br[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, k, x;
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) {
    cin >> x;
    adj[x].push_back(i);
  }
  for (long long int i = 1; i <= 5050; i++) {
    if (adj[i].size() > k) {
      cout << "NO" << endl;
      return 0;
    }
  }
  solve(n, k);
  return 0;
}
