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;
set<int> S[5010];
int allcnt;
int n, k;
int ans[5010];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    S[x].insert(i);
  }
  allcnt = n;
  int mxcnt = 0;
  for (int i = 1; i <= 5000; i++) mxcnt = max(mxcnt, (int)S[i].size());
  if (mxcnt > k) return puts("NO"), 0;
  mxcnt = 0;
  puts("YES");
  while (allcnt) {
    mxcnt++;
    for (int i = 1; i <= 5000; i++)
      if (S[i].size()) {
        ans[*S[i].begin()] = mxcnt;
        S[i].erase(S[i].begin());
        allcnt--;
      }
  }
  while (mxcnt < k) {
    set<int> s;
    mxcnt++;
    for (int i = 1; i <= n; i++) {
      if (s.count(ans[i]) == 0)
        s.insert(ans[i]);
      else {
        ans[i] = mxcnt;
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
