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 = 2e5 + 10;
bool is_Prime(long long a) {
  for (long long i = 2; i * i <= a; ++i)
    if (a % i == 0) return false;
  return true;
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.second < b.second);
}
long long fact(long long n) {
  vector<long long> v;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  return v.size();
}
void C(long long n, long long r) {
  vector<vector<long long>> c(n + 1, vector<long long>(r + 1, 0));
  for (long long i = 0; i < n + 1; i++) {
    for (long long j = 0; j <= i && j <= r; j++)
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
}
bool is_pl(string s, int n) {
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) return false;
  }
  return true;
}
void solve() {
  int n, k;
  cin >> n >> k;
  int cnt[5001] = {0};
  vector<pair<int, int>> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first;
    cnt[v[i].first]++;
    v[i].second = (int)i;
  }
  for (long long i = 0; i < 5001; i++) {
    if (cnt[i] > k) {
      cout << "NO\n";
      return;
    }
  }
  sort(v.begin(), v.end());
  vector<int> ans(n);
  int a = 1;
  for (long long i = 0; i < n; i++) {
    ans[v[i].second] = a;
    a++;
    if (a > k) a = 1;
  }
  cout << "YES\n";
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
