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;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int MaxN = 1e5 + 105;
int m;
string S;
vector<int> pos[56];
set<pair<int, pair<int, int> > > s;
int ans[56];
vector<pair<int, pair<int, int> > > toAdd, toDelete;
void make(int L, int R, char q) {
  toDelete.push_back(make_pair(L - R - 1, make_pair(L, R)));
  int h = (int)(q - 'a');
  int l = 0;
  int r = pos[h].size() - 1;
  int ans1 = pos[h].size();
  while (r >= l) {
    int mid = (l + r) / 2;
    if (pos[h][mid] >= L) {
      ans1 = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  int lastPos = L - 1;
  vector<pair<int, int> > res;
  for (int i = ans1; i < pos[h].size() && pos[h][i] <= R; ++i) {
    int cur = pos[h][i];
    if ((cur - lastPos > m) ||
        (R - lastPos >= m && (i + 1 == pos[h].size() || pos[h][i + 1] > R))) {
      int j = i - 1;
      if (j >= 0 && pos[h][j] >= L && pos[h][j] != lastPos) {
        res.push_back(make_pair(lastPos + 1, pos[h][j] - 1));
        lastPos = pos[h][j];
        --i;
        ++ans[h];
        continue;
      } else {
        res.push_back(make_pair(lastPos + 1, pos[h][i] - 1));
        lastPos = pos[h][i];
        ++ans[h];
      }
    }
  }
  res.push_back(make_pair(lastPos + 1, R));
  for (int i = 0; i < res.size(); ++i) {
    if (res[i].second - res[i].first + 1 >= m)
      toAdd.push_back(make_pair(-(res[i].second - res[i].first + 1), res[i]));
  }
}
void solve(char q) {
  set<pair<int, pair<int, int> > >::iterator it;
  toDelete.clear();
  toAdd.clear();
  for (set<pair<int, pair<int, int> > >::iterator it = s.begin(); it != s.end();
       ++it) {
    pair<int, pair<int, int> > cur = (*it);
    make(cur.second.first, cur.second.second, q);
  }
  s.clear();
  for (int i = 0; i < toAdd.size(); ++i) s.insert(toAdd[i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> m;
  cin >> S;
  for (int i = 0; i < S.size(); ++i) pos[S[i] - 'a'].push_back(i);
  s.insert(make_pair(-S.size(), make_pair(0, S.size() - 1)));
  for (char q = 'a'; q <= 'z'; ++q) {
    solve(q);
    if (s.empty()) {
      for (char q1 = 'a'; q1 < q; ++q1) {
        for (int i = 0; i < (int)pos[q1 - 'a'].size(); ++i) cout << q1;
      }
      for (int i = 0; i < ans[q - 'a']; ++i) cout << q;
      return 0;
    }
  }
}
