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 lx = 3e5 + 123;
const long long int mod = 1e9 + 7;
void solve() {
  int n, i, j, k, l, f = 0, c, clr = 1, b, y = 0;
  cin >> n >> c;
  int a[n];
  vector<pair<int, int> > v(n);
  for (i = 0; i < n; i++) {
    cin >> k;
    v[i].first = k;
    v[i].second = i;
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n;) {
    l = i;
    k = 0;
    while (i < n && v[i].first == v[l].first) i++, k++;
    if (k > c) {
      f++;
      break;
    }
    b = clr;
    for (j = l; j < i; j++) {
      if (clr > c) {
        y++;
        clr = 1;
      }
      if (clr == c) y++;
      a[v[j].second] = clr++;
    }
  }
  if (f || !y)
    cout << "NO"
         << "\n";
  else {
    cout << "YES"
         << "\n";
    for (i = 0; i < n; i++) cout << a[i] << " ";
    cout << "\n";
  }
}
void init_code() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
