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;
mt19937 rnd(time(0));
bool isPrime(long long a) {
  int i;
  for (i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return 0;
    }
  }
  return (a != 1);
}
bool isPalindrom(string s, int l, int r) {
  for (int i = 0; i < (r - l + 1) / 2; i++) {
    if (s[l + i] != s[r - i]) {
      return 0;
    }
  }
  return 1;
}
template <class type1>
type1 gcd(type1 a, type1 b) {
  type1 t;
  while (b) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
int pos[5005];
int used[5005][5005];
vector<int> inf;
int res[5005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k, i, j, a, b;
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> a;
    inf.push_back(a);
  }
  for (auto &it : pos) {
    it = 1;
  }
  for (i = 0; i < k; i++) {
    res[i] = i + 1;
    used[inf[i]][i + 1] = 1;
  }
  for (; i < n; i++) {
    while (used[inf[i]][pos[inf[i]]] == 1) {
      pos[inf[i]]++;
    }
    if (pos[inf[i]] > k) {
      cout << "NO";
      return 0;
    }
    used[inf[i]][pos[inf[i]]] = 1;
    res[i] = pos[inf[i]];
  }
  cout << "YES\n";
  for (i = 0; i < n; i++) {
    cout << res[i] << " ";
  }
  return 0;
}
