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>
int ex[] = {1, -1, 0, 0};
int wye[] = {0, 0, 1, -1};
using namespace std;
int m, n, a[300005], lol;
char second[300005], an[300005];
vector<int> vt[30];
int check(int cnt) {
  int beg = 0, fin = 0, tmp = 0, tpp = 0, last = 0, pop;
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    tmp = 0;
    fin = (beg + m);
    if (fin >= n && tpp == 0 && flag) {
      tpp++;
      break;
    }
    while (i <= fin && i <= n) {
      if (a[i] == cnt) {
        last = i;
      } else if (a[i]) {
        tmp = i;
      }
      i++;
    }
    if (i > n)
      break;
    else if (tmp == 0 && last > beg) {
      tpp++;
      tmp = last;
    } else if (tmp == 0 && last <= beg)
      return 0;
    i = fin;
    flag = false;
    beg = tmp;
  }
  return tpp;
}
void mark(int cnt) {
  int siz = vt[cnt].size(), tmp;
  for (int i = 0; i < siz; i++) {
    tmp = vt[cnt][i];
    a[tmp] = cnt;
    an[lol] = second[tmp - 1];
    lol++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> m >> second;
  n = strlen(second);
  for (int i = 0; i < n; i++) {
    vt[(int)(second[i] - 'a') + 1].push_back(i + 1);
  }
  for (int i = 'a', cnt = 1; i <= 'z'; i++, cnt++) {
    mark(cnt);
    int tmp = check(cnt);
    if (tmp) {
      tmp = lol - vt[cnt].size() + tmp;
      an[tmp] = '\0';
      break;
    }
  }
  cout << an << endl;
  return 0;
}
