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 ppr = 257;
const long long INF = 1e18;
const int inf = 1e9;
const int mod = 1e9 + 7;
const int N = 1e6 + 123;
const long double pi = 3.141592653589793238462643;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int n, k, a[N], c[N], last[N];
set<pair<int, int> > st;
pair<int, int> p[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[i] = make_pair(a[i], i);
  }
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= k; i++) {
    c[p[i].second] = i;
    last[i] = p[i].first;
  }
  for (int i = k + 1; i <= n; i++) st.insert(p[i]);
  set<pair<int, int> >::iterator it;
  for (int i = 1; i <= k; i++) {
    vector<pair<int, int> > v;
    for (it = st.begin(); it != st.end(); it++) {
      if ((*it).first > last[i]) {
        v.push_back((*it));
        last[i] = (*it).first;
        c[(*it).second] = i;
      }
    }
    for (int j = 0; j < v.size(); j++) st.erase(v[j]);
  }
  for (int i = 1; i <= n; i++)
    if (!c[i]) return cout << "NO", 0;
  cout << "YES\n";
  for (int i = 1; i <= n; i++) cout << c[i] << ' ';
  return 0;
}
