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>
long long read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) res = res * 10 + ch - 48, ch = getchar();
  return res * flag;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
const long long N = 5000 + 5;
long long n, k, a[N], c[N];
std::vector<long long> g[N];
void solution() {
  n = read(), k = read();
  long long tot = 0;
  for (long long i = (1); i <= (n); ++i) a[i] = read(), g[a[i]].push_back(i);
  for (long long i = (1); i <= (N - 1); ++i) {
    if (g[i].size() > k) {
      puts("NO");
      return;
    }
    for (auto v : g[i]) c[v] = ++tot, tot %= k;
  }
  puts("YES");
  for (long long i = (1); i <= (n); ++i) print(c[i]), putchar(' ');
  putchar('\n');
}
signed main() {
  42;
  solution();
}
