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;
int n, k;
int a[5005];
int b[5005];
int buck[5005];
bool qflag;
int main() {
  while (cin >> n >> k) {
    memset(buck, 0, sizeof(buck));
    qflag = false;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = a[i];
      buck[a[i]]++;
    }
    for (int i = 1; i <= 5000; i++) {
      if (buck[i] > k) {
        qflag = true;
        break;
      }
    }
    if (qflag) {
      cout << "NO" << endl;
      continue;
    }
    sort(b, b + n);
    buck[b[0]] = 1;
    for (int i = 0; i < n; i++) {
      if (i < (n - 1) && b[i + 1] != b[i]) {
        buck[b[i + 1]] = (i + 1) % k + 1;
      }
    }
    for (int i = 0; i < n; i++) {
      int tres = buck[a[i]];
      buck[a[i]]++;
      if (buck[a[i]] > k) buck[a[i]] -= k;
      a[i] = tres;
    }
    cout << "YES" << endl;
    cout << a[0];
    for (int i = 1; i < n; i++) {
      cout << ' ' << a[i];
    }
    cout << endl;
  }
  return 0;
}
