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>
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
inline int ones(int n) {
  int res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
template <class T>
void chmax(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void chmin(T& a, const T& b) {
  a = min(a, b);
}
long long gcd(long long a, long long b) {
  if (a == 0 || b == 0) return 0;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long powermod(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long mulmod(long long a, long long b) {
  long long res = 0;
  a %= 1000000007;
  while (b) {
    if (b & 1) res = (res + a) % 1000000007;
    a = (2 * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i += 6) {
    if ((n % i == 0) || (n % (i + 2) == 0)) return false;
  }
  return true;
}
long double dist(long long x1, long long y1, long long x2, long long y2) {
  return (long double)sqrt(
      (long double)((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)));
}
long long squaredist(long long x1, long long y1, long long x2, long long y2) {
  return ((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long nCr(long long n, long long r) {
  if (r == 0) return 1;
  return (n * nCr(n - 1, r - 1)) / r;
}
long long countDivisors(long long n) {
  long long cnt = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int a[5001];
  int h[5001] = {0};
  int h2[5001] = {0};
  int sc[5001] = {0};
  int uc = 0;
  int maxi = INT_MIN;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    if (!(h[a[i]])) uc++;
    h[a[i]]++;
    maxi = max(maxi, h[a[i]]);
  }
  if (maxi > k) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  vector<pair<int, int>> v1;
  vector<pair<int, int>> v2;
  for (int i = 0; i < (n); i++) {
    v1.push_back(make_pair(a[i], i));
  }
  sort(v1.begin(), v1.end());
  int count = 0;
  for (int i = 0; i < (n); i++) {
    v2.push_back(make_pair(v1[i].second, count % k + 1));
    count++;
  }
  sort(v2.begin(), v2.end());
  for (int i = 0; i < (n); i++) {
    cout << v2[i].second << " ";
  }
  return 0;
}
