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;
long long expo(long long x, long long n, long long m) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return expo((x * x) % m, n / 2, m);
  else
    return (x * expo((x * x) % m, (n - 1) / 2, m)) % m;
}
long long fun(long long k) {
  long long ans = 1;
  for (int i = 0; i < 3; i++) {
    ans *= (k - i);
  }
  for (int i = 1; i <= 3; i++) {
    ans = ans / i;
  }
  return ans;
}
void solve() {
  int n, k;
  cin >> n >> k;
  int count[5010];
  memset(count, 0, sizeof(count));
  ;
  int arr[n + 1];
  int maxx = 1;
  int colour = 1;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    arr[i] = t;
    count[t]++;
    if (count[t] > maxx) maxx = count[t];
  }
  if (maxx > k)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    int ans[n + 1];
    for (int i = 1; i <= 5000; i++) {
      for (int j = 1; j <= n; j++) {
        if (arr[j] == i) {
          ans[j] = colour;
          colour++;
          if (colour > k) colour = 1;
        }
      }
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) solve();
  return 0;
}
