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>
const double PI = 2.0 * acos(0.0);
const double EPS = 1e-9;
using namespace std;
int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dc[] = {0, 1, 0, -1, -1, 1, -1, 1};
int V, E;
int T;
int N, M, K;
string S;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> N >> K;
  int ans[5010];
  int i, j;
  map<int, vector<int>> m;
  for (i = 0; i <= N - 1; i++) cin >> j, m[j].push_back(i);
  int col = 0;
  for (auto x : m) {
    if (x.second.size() > K) {
      cout << "NO" << endl;
      return 0;
    } else {
      for (j = 0; j <= x.second.size() - 1; j++) {
        ans[x.second[j]] = col % K + 1;
        col++;
      }
    }
  }
  cout << "YES" << endl;
  for (i = 0; i <= N - 1; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}
