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 fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
struct st {
  int val;
  int col;
  int idx;
} arr[50100];
vector<int> G[50005];
bool cmp1(st lhs, st rhs) { return lhs.val < rhs.val; }
bool cmp2(st lhs, st rhs) { return lhs.idx < rhs.idx; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i].val);
    arr[i].idx = i;
  }
  int valid = 1;
  sort(arr, arr + n, &cmp1);
  for (int i = 0; i < n; i++) {
    if (i + 1 < n && (arr[i].val == arr[i + 1].val)) {
      int st = i;
      while (i + 1 < n && (arr[i].val == arr[i + 1].val)) {
        i++;
      }
      int tot = i - st + 1;
      if (tot > k) valid = 0;
      int cnt = 1;
      for (int j = st; j <= i; j++) {
        arr[j].col = cnt;
        G[cnt].push_back(j);
        cnt++;
      }
    } else {
      arr[i].col = 1;
      G[1].push_back(i);
    }
  }
  for (int i = 1; i <= k; i++) {
    if (G[i].size() == 0) {
      bool pos = 0;
      for (int j = 1; j <= k; j++) {
        if (G[j].size() > 1 && G[i].size() == 0) {
          int idd = G[j][G[j].size() - 1];
          arr[idd].col = i;
          G[j].pop_back();
          G[i].push_back(idd);
          pos = 1;
        }
      }
      if (!pos) valid = 0;
    }
  }
  sort(arr, arr + n, &cmp2);
  if (valid) {
    printf("YES\n");
    for (int i = 0; i < n; i++) {
      printf("%d ", arr[i].col);
    }
    printf("\n");
  } else
    printf("NO\n");
}
