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;
bool visit[2010][2010] = {};
vector<long long> vec[100001];
long long b[100001] = {};
struct s {
  long long s1, s2;
} a[10000];
bool srt(const s &lhs, const s &rhs) {
  if (lhs.s1 < rhs.s1)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(false);
  long long n, m, x, k;
  cin >> n >> k;
  map<long long, long long> mp;
  map<long long, long long> mlp;
  bool f = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i].s1;
    a[i].s2 = i;
    mlp[i] = 0;
    if (mp.find(a[i].s1) == mp.end()) {
      mp[a[i].s1] = 1;
    } else {
      mp[a[i].s1]++;
    }
    if (mp[a[i].s1] > k) {
      f = 1;
    }
  }
  if (f == 1 || n < k) {
    cout << "NO\n";
    return 0;
  }
  int r = 1;
  stable_sort(a, a + n, srt);
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (a[j].s1 == a[i].s1 && mlp[i] == 0) {
        mlp[i] = 1;
        b[a[i].s2] = r;
        r++;
        if (r > k) r = 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
}
