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>
const long double PI = acos(-1);
using namespace std;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
bool vis[5123][5123] = {0};
bool used[5123] = {0};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long v[n];
  long long ans[n];
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long prior = 1;
  for (long long i = 0; i < n; i++) {
    long long use = -1;
    while (used[prior] && prior <= k) prior++;
    for (long long j = 1; j <= k; j++) {
      if (vis[v[i]][j]) continue;
      use = j;
      break;
    }
    if (use == -1) {
      cout << "NO" << '\n';
      return 0;
    }
    if (!vis[v[i]][prior] && prior <= k) {
      use = prior;
      prior++;
    }
    used[use] = true;
    vis[v[i]][use] = true;
    ans[i] = use;
  }
  if (prior < k) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  cout << '\n';
}
