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;
ifstream fi("test.txt");
ofstream fo("out.txt");
const long long MOD = 1e9 + 7;
const long long base = 269;
const int ooi = 1 << 30;
const int N = 1e5 + 5;
long long n, m, q, k, pos, rs, t, cnt, rs1, check;
long long maxx = -1e9, minn = 1e9;
long long l, r, d;
int a[N], mark[N], color = 1;
int usecolor[N], ans[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (long long(i) = 0; (i) < (n); (i)++) {
    cin >> a[i];
    mark[a[i]]++;
  }
  for (long long(i) = 0; (i) < (5005); (i)++) {
    if (mark[i]) {
      if (mark[i] <= k) {
        for (long long(j) = 0; (j) < (n); (j)++) {
          if (a[j] == i) {
            ans[j] = color;
            usecolor[color]++;
            ++color;
            if (color > k) color = 1;
          }
        }
      } else
        return cout << "NO", 0;
    }
  }
  for (long long(i) = 1; (i) <= (k); (i)++)
    if (!usecolor[i]) return cout << "NO", 0;
  cout << "YES\n";
  for (long long(i) = 0; (i) < (n); (i)++) cout << ans[i] << " ";
}
