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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD1 = 1e9 + 7, MOD2 = 1e9 + 9;
const int N = 5e3 + 10;
int n, k, a[N], mark[N], c[N], beg[N];
set<int> uc[N];
vector<int> idx[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int cur = 1;
  c[0] = cur, uc[a[0]].insert(cur), mark[cur] = 1, idx[cur].push_back(0);
  for (int i = 1; i < n; i++) {
    if (a[i] != a[i - 1] && uc[a[i]].find(cur) == uc[a[i]].end())
      c[i] = cur, uc[a[i]].insert(cur);
    else {
      if (uc[a[i]].size() == k)
        printf("NO\n"), exit(0);
      else {
        for (int j = 1; j <= k; j++) {
          if (uc[a[i]].find(j) == uc[a[i]].end()) {
            uc[a[i]].insert(j), cur = j;
            break;
          }
        }
        c[i] = cur, uc[a[i]].insert(cur);
      }
    }
    mark[cur] = 1, idx[cur].push_back(i);
  }
  int j = -1;
  for (int i = 1; i <= k; i++) {
    if (!mark[i]) {
      j = i;
      break;
    }
  }
  if (j != -1) {
    bool ok = false;
    int cur = k, cc = 1;
    while (cur >= j) {
      while ((beg[cc] >= idx[cc].size() - 1 or idx[cc].size() == 1) && cc <= k)
        cc++;
      ;
      if (cc > k) printf("NO\n"), exit(0);
      int lst = beg[cc] - 1;
      for (int i = beg[cc]; i < (int)idx[cc].size() - 1 && cur >= j; i++)
        c[idx[cc][i]] = cur--, lst = i;
      beg[cc] = lst + 1;
    }
  }
  puts("YES");
  for (int i = 0; i < n; i++) printf("%d ", c[i]);
}
