You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ n) from the string s. Polycarp uses the following algorithm k times:

  * if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; 
  * if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; 
  * ... 
  * remove the leftmost occurrence of the letter 'z' and stop the algorithm. 



This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers n and k (1 ≤ k ≤ n ≤ 4 ⋅ 10^5) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string s consisting of n lowercase Latin letters.

Output

Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.

If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).

Examples

Input

15 3
cccaabababaccbc


Output

cccbbabaccbc


Input

15 9
cccaabababaccbc


Output

cccccc


Input

1 1
u


Output
#include <bits/stdc++.h>
using namespace std;
template <typename T, size_t size>
void print(const T (&array)[size], size_t n) {
  cerr << "[";
  for (size_t i = 0; i < n; ++i) {
    if (i != 0) {
      cerr << ", ";
    }
    cerr << array[i];
  }
  cerr << "]";
  printf("\n");
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
long long int gcd(long long int A, long long int B) {
  if (B == 0)
    return A;
  else
    return gcd(B, A % B);
}
long long int mod_exponentiation(long long int base, long long int n,
                                 long long int m) {
  if (base == 0) return 0;
  if (n == 0) return 1 % m;
  long long int half = mod_exponentiation(base, n / 2, m);
  if (n % 2 == 0)
    return (half % m * half % m) % m;
  else
    return ((half % m * half % m) % m * (base % m)) % m;
}
int a[26];
int cnt[26];
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  for (int i = 0; i < (int)s.size(); i++) {
    int x = s[i] - 'a';
    a[x]++;
  }
  int pref[26];
  pref[0] = a[0];
  for (int i = 1; i < 25; i++) {
    pref[i] = pref[i - 1] + a[i];
  }
  int idx = lower_bound(pref, pref + 26, k) - pref;
  int op = 0;
  while (op != k) {
    for (int i = 0; i < 26; i++) {
      if (a[i] != 0) {
        a[i]--;
        break;
      }
    }
    op++;
  }
  string c = "";
  stack<char> st;
  for (int i = n - 1; i >= 0; i--) {
    int x = s[i] - 'a';
    if (cnt[x] >= a[x]) {
      continue;
    }
    cnt[x]++;
    st.push(s[i]);
  }
  while (!st.empty()) {
    cout << st.top();
    st.pop();
  }
}
