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;
const long long maxn = 5050;
long long a[maxn];
long long c[maxn];
long long tag[maxn];
long long number[maxn];
int main() {
  long long i, j, k, n, m;
  while (scanf("%lld%lld", &n, &k) != EOF) {
    memset(tag, 0, sizeof(tag));
    for (i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      tag[a[i]]++;
    }
    for (i = 1; i <= 5000; i++) number[i] = 1;
    bool flag = true;
    for (i = 1; i <= 5000; i++)
      if (tag[i] > k) {
        flag = false;
        break;
      }
    if (!flag)
      printf("NO\n");
    else {
      long long cnt = 1;
      for (i = 1; i <= 5000; i++) {
        for (j = 1; j <= 5000; j++)
          if (a[j] == i) {
            c[j] = cnt;
            cnt++;
            if (cnt > k) cnt = 1;
          }
      }
      printf("YES\n");
      for (i = 1; i <= n; i++) printf("%lld ", c[i]);
      printf("\n");
    }
  }
}
