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 int N = 1000001;
const int oo = 1e9;
const int MOD = 1000000007;
int n, a, k, ar[5020];
bool flag;
long long sum = 0;
int f[5020];
vector<vector<int>> v(50001);
int main() {
  cin >> n >> k;
  v.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a);
    f[a]++;
    if (f[a] > k) flag = 1;
    ar[i] = f[a];
    v[a].push_back(i);
  }
  if (flag) {
    puts("NO");
    return 0;
  }
  puts("YES");
  int c = k;
  for (int i = 1; i <= 5000; ++i) {
    for (int j = 0; j < v[i].size(); j++) {
      ar[v[i][j]] = c;
      c--;
      if (c == 0) c = k;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ar[i] << " ";
  }
  return 0;
}
