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 c[26] = {0};
const int maxm = 15000005;
bool prime[maxm];
void seive() {
  memset(prime, true, sizeof(prime));
  long long p;
  for (p = 2; p * p < maxm; p++) {
    if (prime[p]) {
      for (long long i = 2 * p; i < maxm; i += p) prime[i] = false;
    }
  }
  prime[0] = false;
  prime[1] = false;
  return;
}
void solve() {
  map<long long, long long> m;
  long long n, x, z, mx = INT_MIN, i = 0, j = 0, k = 0, first = 1, one = 0,
                     uniq = 0;
  cin >> n >> k;
  if (k > n) {
    cout << "NO";
    return;
  }
  vector<pair<long long, long long> > v(n);
  vector<long long> ans(n), color(k + 1, 0);
  for (i = 0; i < n; i++) {
    cin >> x;
    v[i] = {x, i};
    m[x]++;
  }
  for (auto it : m) {
    if (it.second > k) {
      cout << "NO";
      return;
    }
  }
  sort(v.begin(), v.end());
  i = 0;
  while (i < n) {
    x = 1;
    j = i;
    first = 0;
    while (j < n && v[i].first == v[j].first) {
      color[x]++;
      z = v[j].second;
      ans[z] = x;
      x++;
      j++;
    }
    i = j;
  }
  for (i = 1; i <= k; i++) {
    if (color[i] == 0) break;
  }
  z = i;
  for (i = 0; i < n && z <= k; i++) {
    if (color[ans[i]] > 1) {
      color[ans[i]]--;
      ans[i] = z++;
    }
  }
  cout << "YES\n";
  for (i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  return;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  solve();
}
