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;
vector<long long> graph[100002];
vector<long long> respect(100002);
set<long long> s;
vector<bool> vis(100002, false);
void dfs(long long p) {
  long long val = respect[p];
  vis[p] = true;
  for (auto it : graph[p]) {
    if (vis[it]) continue;
    val *= respect[it];
    dfs(it);
  }
  if (val) s.insert(p);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  map<long long, long long> cnt;
  vector<pair<long long, pair<long long, long long>>> v;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    v.push_back(make_pair(x, make_pair(i, -1)));
    cnt[x]++;
    if (cnt[x] > k) {
      cout << "NO"
           << "\n";
      return;
    }
  }
  sort(v.begin(), v.end(),
       [](pair<long long, pair<long long, long long>> &x,
          pair<long long, pair<long long, long long>> &y) {
         return x.first < y.first;
       });
  for (long long i = 0, temp = 1; i < v.size(); i++) {
    v[i].second.second = temp;
    temp++;
    if (temp > k) temp = 1;
  }
  sort(v.begin(), v.end(),
       [](pair<long long, pair<long long, long long>> &x,
          pair<long long, pair<long long, long long>> &y) {
         return x.second.first < y.second.first;
       });
  cout << "YES"
       << "\n";
  for (auto i : v) {
    cout << i.second.second << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  int kse = 1;
  while (kse--) {
    solve();
  }
}
