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;
int n, m;
string s;
set<pair<int, int> > st[100];
set<int> remain;
int f[300];
int in(int x, int a, int b) { return a <= x && x <= b; }
int ok(int x) {
  int ret = 1;
  int pos = *remain.begin();
  auto cur = st[x].begin();
  for (auto i : remain) {
    if (in(i, cur->first, cur->second)) continue;
    for (; cur != st[x].end() && !in(i, cur->first, cur->second); cur++)
      ;
    if (cur == st[x].end()) return -1;
    if (!in(pos, cur->first, cur->second)) {
      ret++;
      pos = i;
    }
  }
  return ret;
}
int main() {
  memset(f, 0, sizeof(f));
  cin >> m >> s;
  n = s.size();
  for (int i = (0); i < (n); i++) f[s[i]]++;
  int lo = 0, hi = 0;
  for (int i = (0); i < (n); i++) {
    st[s[i] - 'a'].insert(pair<int, int>(lo, hi));
    hi++;
    lo = max(lo, hi - m + 1);
  }
  for (int i = (0); i <= (n - m); i++) remain.insert(i);
  string ans = "";
  for (int i = (0); i <= ('z' - 'a'); i++) {
    if (st[i].empty()) continue;
    int tmp = ok(i);
    char c = i + 'a';
    if (tmp == -1) {
      for (auto p : st[i]) {
        int x = p.first, y = p.second;
        for (auto it = remain.lower_bound(x); it != remain.end();
             it = remain.lower_bound(x)) {
          if (*it <= y)
            remain.erase(it);
          else
            break;
        }
      }
      ans += string(f[c], c);
    } else {
      ans += string(tmp, c);
      break;
    }
  }
  cout << ans << '\n';
  return 0;
}
