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 MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 2 * 1e18;
const long double PI = 3.14159265358979323846;
long long a[200005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  bool x = 0;
  map<pair<long long, long long>, long long> ans;
  map<long long, long long> m;
  vector<pair<long long, long long> > v;
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(make_pair(a[i], i));
    m[a[i]]++;
    if (m[a[i]] > k) x = 1;
  }
  if (x) {
    cout << "NO";
    return 0;
  }
  sort(v.begin(), v.end());
  long long cnt = 1;
  for (long long i = 0; i < n; i++) {
    ans[v[i]] = cnt;
    cnt++;
    if (cnt > k) cnt = 1;
  }
  cout << "YES\n";
  for (long long i = 0; i < n; i++) {
    cout << ans[make_pair(a[i], i)] << " ";
  }
}
