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 sqr(long long x) { return x * x; }
long double sqrd(long double x) { return x * x; }
mt19937 Random((int)time(0));
long long rnd(long long x) { return Random() % x; }
const long long INF = (long long)1e9 + 282;
const long long MOD = (long long)1e9 + 7;
const long double PI = acosl(-1.0);
const int N = 5010;
int cnt[N], ans[N];
vector<int> w[N];
void mane() {
  int n, m;
  cin >> n >> m;
  int res = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    ans[i] = ++cnt[a];
    if (cnt[a] > m) {
      cout << "NO";
      return;
    }
    res = max(res, cnt[a]);
    w[cnt[a]].push_back(i);
  }
  if (res < m) {
    for (int i = 0; i < N; i++) {
      for (int j = 1; j < (int)(w[i]).size() && res < m; j++) {
        ans[w[i][j]] = ++res;
      }
    }
  }
  if (res < m) {
    cout << "NO";
    return;
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios_base::fixed, ios_base::floatfield);
  cout.precision(12);
  mane();
}
