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;
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool sortByVal(const pair<double, double> &a, const pair<double, double> &b) {
  if (a.first == b.first) {
    return (a.second > b.second);
  } else
    return (a.first < b.first);
}
long long Powb(long long b, long long n) {
  if (n == 0) return 1LL;
  if (n == 1) return b;
  long long temp = Powb(b, n / 2);
  if (n % 2 == 0) {
    return (temp * temp) % 1000000007;
  } else {
    return (b * ((temp * temp) % 1000000007)) % 1000000007;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  while (t--) {
    long long k = 0, n, i = 0, j = 0, c = 0, d = 0, e = 0, f = 0, h = 0,
              maxi = 0, sum = 0;
    cin >> n >> k;
    vector<long long> a(n);
    map<long long, vector<long long> > m;
    vector<long long> b, p(n);
    for (i = 0; i < n; i++) {
      cin >> a[i];
      m[a[i]].push_back(i);
    }
    c = 1;
    for (auto x : m) {
      b = x.second;
      if (b.size() > k) {
        f++;
        break;
      } else {
        for (i = 0; i < b.size(); i++) {
          if (c > k) c = 1;
          p[b[i]] = c;
          c++;
        }
      }
    }
    if (f != 0)
      cout << "NO"
           << "\n";
    else {
      cout << "YES"
           << "\n";
      {
        for (int i = 0; i < n; i++) {
          cout << p[i] << " ";
        }
        cout << "\n";
      };
    }
  }
}
