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;
template <class A, class B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  return out << "(" << a.first << "," << a.second << ")";
}
template <class A>
ostream &operator<<(ostream &out, const vector<A> &a) {
  for (const A &it : a) out << it << " ";
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  return in >> a.first >> a.second;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &i : a) in >> i;
  return in;
}
ifstream cinn("in.txt");
ofstream coutt("out.txt");
long long poww(const long long &a, long long b,
               const long long &m = 1000000007) {
  if (b == 0) return 1;
  long long first = poww(a, b / 2, m);
  first = first * first % m;
  if (b & 1) first = first * a % m;
  return first;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long _sum(long long n) { return (n * (n + 1)) / 2; }
bool isPalindrome(string str) {
  long long l = 0;
  long long h = str.length() - 1;
  while (h > l) {
    if (str[l++] != str[h--]) {
      return false;
    }
  }
  return true;
}
long long fact(long long n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
long long distSq(pair<long long, long long> p, pair<long long, long long> q) {
  return (p.first - q.first) * (p.first - q.first) +
         (p.second - q.second) * (p.second - q.second);
}
vector<long long> adj[(long long)1e5 + 5];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, n, c, h, b, m, k, f, f2, t2, t3, i, j, ans, mnt, mxt, r, j1, j2;
  cin >> n >> k;
  vector<long long> v(n);
  cin >> v;
  vector<long long> kk = v;
  sort((kk).begin(), (kk).end());
  adj[(long long)1].push_back(kk[0]);
  j = 1;
  f = 0;
  for (i = 1; i < n; i++) {
    if (kk[i] == kk[i - 1]) {
      j++;
      if (j <= k) {
        adj[j].push_back(kk[i]);
      } else {
        f = 1;
        break;
      }
    } else {
      j = 1;
      adj[j].push_back(kk[i]);
    }
  }
  if (f) {
    cout << "NO" << '\n';
    return 0;
  }
  cout << "YES" << '\n';
  vector<long long> col(n, 0);
  map<long long, long long> sele, oc;
  for (i = 1; i < k + 1; i++) {
    map<long long, long long> mm;
    for (j = 0; j < n; j++) {
      if (col[j] == 0 && !mm[v[j]]) {
        sele[i] = 1;
        col[j] = i;
        mm[v[j]] = 1;
        oc[i]++;
      }
    }
  }
  for (i = 1; i < k + 1; i++) {
    if (!sele[i]) {
      for (j = 0; j < n; j++) {
        if (oc[col[j]] > 1) {
          oc[col[j]]--;
          col[j] = i;
          oc[col[j]]++;
          break;
        }
      }
    }
  }
  cout << col << '\n';
}
