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 = 5e3 + 10;
const int MOD = 1e9 + 7;
const double PI = acos(-1);
int a[N];
int c[N];
vector<int> v[N];
bool usedColor[N];
int main() {
  int n, k;
  cin >> n >> k;
  int mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
    v[a[i]].push_back(i);
  }
  int colorID = 0;
  for (int i = 1; i <= mx; i++) {
    if (v[i].size() > k) {
      cout << "NO";
      return 0;
    }
    for (int j = 0; j < v[i].size(); j++) {
      c[v[i][j]] = colorID++;
      usedColor[colorID] = 1;
      colorID %= k;
    }
  }
  for (int i = 1; i <= k; i++) {
    if (!usedColor[i]) {
      cout << "NO";
      return 0;
    }
  }
  puts("YES");
  for (int i = 0; i < n; i++) {
    cout << c[i] + 1 << ' ';
  }
  return 0;
}
