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>
const int lim = 1e6;
int aux[lim], cubeta[lim], arr[lim];
int n, m, mayor, sum = 0, esp;
void can() {
  int temporal = m;
  while (temporal > 0) {
    temporal -= cubeta[sum];
    if (temporal < 0) {
      esp = sum;
    }
    sum++;
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    int temp;
    scanf("%d", &temp);
    cubeta[temp]++;
    arr[i] = temp;
    if (cubeta[temp] > m) {
      printf("NO\n");
      return 0;
    }
  }
  can();
  int mayor = m, otro = 1;
  printf("YES\n");
  for (int i = 0; i < n; i++) {
    if (arr[i] > sum - 1) {
      aux[arr[i]]++;
      printf("%d ", aux[arr[i]]);
    } else if (arr[i] == esp) {
      printf("%d ", otro++);
    } else {
      printf("%d ", mayor--);
    }
  }
  printf("\n");
  return 0;
}
