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;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long int bst(long long int a[], long long int item, long long int index,
                  long long int n) {
  long long int b = index, e = n, mid;
  mid = (b + e) / 2;
  while (a[mid] != item && b <= e) {
    if (a[mid] < item)
      b = mid + 1;
    else
      e = mid - 1;
    mid = (b + e) / 2;
  }
  return mid;
}
long long int max3(long long int x, long long int y, long long int z) {
  return max(max(x, y), z);
}
long long int min3(long long int x, long long int y, long long int z) {
  return min(min(x, y), z);
}
string cir_shift(string s) {
  string s1;
  long long int len;
  len = s.size();
  s1 = s.substr(1, len - 1);
  s1.push_back(s[0]);
  return s1;
}
template <typename Type>
void swap1(Type &a, Type &b) {
  Type temp;
  temp = a;
  a = b;
  b = temp;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long int n, k;
  cin >> n >> k;
  long long int ar[n], ar1[n];
  unordered_map<int, long long int> mp;
  long long int c = 0;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    mp[ar[i]]++;
    c = max(c, mp[ar[i]]);
  }
  if (k < c || k > n) {
    cout << "NO\n";
  } else {
    cout << "YES" << endl;
    int p = 0;
    for (int i = 0; i < n; i++) {
      if (mp[ar[i]] != 0) {
        for (int j = i; j < n; j++) {
          if (ar[j] == ar[i]) {
            ar1[j] = p % k + 1;
            p++;
          }
        }
        mp[ar[i]] = 0;
      }
    }
    for (int i = 0; i < n; i++) cout << ar1[i] << " ";
    cout << endl;
  }
}
