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 pre[5005][5005];
int main() {
  int n, k;
  bool flag = true;
  scanf("%d%d", &n, &k);
  map<int, int> m;
  for (int i = 1; i <= n; i++) {
    int yy;
    scanf("%d", &yy);
    m[yy]++;
    pre[yy][m[yy]] = i;
    if (m[yy] > k) flag = false;
  }
  int xx = 0;
  bool flag1 = false;
  int p[5005];
  for (int i = 1; i <= 5000; i++) {
    for (int j = 1; j <= m[i]; j++) {
      p[pre[i][j]] = ++xx;
      if (xx == k) {
        flag1 = true;
        xx = 0;
      }
    }
  }
  if (flag == false || flag1 == false)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; i++) {
      cout << p[i] << " ";
    }
  }
  return 0;
}
