You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Examples

Input

3
cbabc


Output

a


Input

2
abcab


Output

aab


Input

3
bcabcbaccba


Output

aaabb

Note

In the first sample, one can choose the subsequence {3} and form a string "a".

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <class T>
inline bool Min(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
inline bool Max(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
inline int ni() {
  int a;
  scanf("%d", &a);
  return a;
}
inline ll nl() {
  ll a;
  scanf("%lld", &a);
  return a;
}
inline double nd() {
  double a;
  scanf("%lf", &a);
  return a;
}
inline void pi(int a) { printf("%d ", a); }
inline void pl(ll a) { printf("%lld ", a); }
inline void pd(double a) { printf("%.12lf ", a); }
void solve() {
  int m;
  string str;
  cin >> m >> str;
  int n = int(str.size()), r = 0;
  char maxi = 'a';
  multiset<char> s;
  vector<int> dp(n);
  for (int i = 0; i < (int)(n); ++i) {
    while (r < n and r < (i + m)) s.insert(str[r++]);
    if (n - i >= m) Max(maxi, *s.begin());
    s.erase(s.find(str[i]));
  }
  set<pair<char, int>> ms;
  r = 0;
  int last = -m;
  int cnt = 0;
  for (int i = 0; i < (int)(n); ++i) {
    while (r < n and r < (i + m)) ms.insert({str[r], -r}), r++;
    if (ms.begin()->first == maxi and n - i >= m) {
      int pos = -ms.begin()->second;
      if (last < i) last = pos, cnt++;
    }
    ms.erase({str[i], -i});
  }
  string ans;
  for (int i = 0; i < (int)(cnt); ++i) ans.push_back(maxi);
  for (int i = 0; i < (int)(n); ++i)
    if (str[i] < maxi) ans.push_back(str[i]);
  sort((ans).begin(), (ans).end());
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
