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;
struct node {
  int v, id;
} a[5010];
int tot = 1, T, col[5010], check[5010], cnt[5010];
bool mycmp(node x, node y) { return x.v < y.v; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].v), a[i].id = i;
  sort(a + 1, a + n + 1, mycmp);
  for (int i = 1; i <= n; ++i) cnt[a[i].v]++, tot = max(tot, cnt[a[i].v]);
  if (tot > k) {
    puts("NO");
    return 0;
  }
  tot = 0;
  for (int i = 1; i <= n; ++i) col[a[i].id] = (++(tot %= k));
  for (int i = 1; i <= n; ++i) T = max(T, col[a[i].id]);
  if (T < k)
    puts("NO");
  else {
    puts("YES");
    for (int i = 1; i <= n; ++i) printf("%d ", col[i]);
  }
  return 0;
}
