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>
const int INF = 0x3f3f3f3f;
const int maxn = 405;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const double eps = 1e-7;
using namespace std;
int n, k, num[5005];
struct node {
  int v;
  int pos;
  int ans;
} e[5005];
bool cmp1(node a, node b) { return a.v < b.v; }
bool cmp2(node a, node b) { return a.pos < b.pos; }
int main() {
  scanf("%d", &(n)), scanf("%d", &(k));
  bool flag = 1;
  for (int i = 0; i <= n - 1; ++i) {
    scanf("%d", &(e[i].v));
    e[i].pos = i;
    num[e[i].v]++;
    if (num[e[i].v] > k) flag = 0;
  }
  if (!flag) return puts("NO"), 0;
  sort(e, e + n, cmp1);
  puts("YES");
  int cnt = 0;
  for (int i = 0; i <= n - 1; ++i) {
    e[i].ans = 1 + (cnt % k);
    cnt++;
  }
  sort(e, e + n, cmp2);
  for (int i = 0; i <= n - 1; ++i) {
    printf("%d%c", e[i].ans, i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
