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 inf = INT_MAX;
long long modulo(long long num, long long MOD = 1000000007) {
  return ((num % MOD) + MOD) % MOD;
}
long long power(long long b, long long e, long long MOD = 1000000007) {
  long long ans = 1;
  while (e) {
    if (e % 2) ans = (ans * b) % MOD;
    b = (b * b) % MOD;
    e /= 2;
  }
  return ans;
}
long long inv(long long num, long long MOD = 1000000007) {
  return power(modulo(num), MOD - 2, MOD);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int f[100005] = {}, k, i, a[100005], n, cnt[100005] = {}, col[100005] = {};
  cin >> n >> k;
  for (i = 0; i < n; i++) cin >> a[i], f[a[i]]++;
  for (i = 0; i < n; i++)
    if (f[a[i]] > k) return cout << "NO", 0;
  cout << "YES\n";
  vector<int> v[100005];
  for (i = 0; i < n; i++) v[a[i]].push_back(i);
  int t = 0;
  for (i = 0; i < 5005; i++) {
    for (int j = 0; j < v[i].size(); j++) {
      col[v[i][j]] = t;
      t++;
      if (t == k) t -= k;
    }
  }
  for (i = 0; i < n; i++) {
    cout << col[i] + 1 << " ";
  }
  return 0;
}
