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 = (int)1e9;
const double eps = 1e-6;
const int mod = inf + 7;
const int N = 100000;
int n, m;
pair<int, int> t[400400];
string second, ans;
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v].first = second[tl] - 'a';
    t[v].second = -tl;
    return;
  }
  int mid = (tl + tr) / 2;
  build(v + v, tl, mid);
  build(v + v + 1, mid + 1, tr);
  t[v] = min(t[v + v], t[v + v + 1]);
}
pair<int, int> get(int v, int tl, int tr, int l, int r) {
  if (l > tr || r < tl) {
    return make_pair(inf, inf);
  }
  if (l <= tl && tr <= r) {
    return t[v];
  }
  int mid = (tl + tr) / 2;
  return min(get(v + v, tl, mid, l, r), get(v + v + 1, mid + 1, tr, l, r));
}
multiset<int> q, st;
int was[10000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> m >> second;
  int n = second.length();
  build(1, 0, n - 1);
  int c = 100;
  for (int i = 0; i < n; ++i) {
    st.insert(second[i] - 'a');
  }
  int cnt = 0;
  for (int i = m - 1; i < n; ++i) {
    int j = i - m + 1;
    pair<int, int> cur = get(1, 0, n - 1, j, i);
    was[cur.first]++;
    j = -cur.second + m;
    if (j > n) {
      cur = get(1, 0, n - 1, i + 1, n - 1);
    }
    i = j - 1;
  }
  sort(second.begin(), second.end());
  for (int i = 0; i < second.length(); ++i) {
    int j = i;
    while (j < n && second[j] == second[i]) {
      j++;
    }
    if (was[second[i] - 'a']) {
      ans += second.substr(i, j - i);
    } else {
      if (j != n) {
        bool ok = 0;
        for (int k = second[i] - 'a'; k <= 'z' - 'a'; ++k) {
          if (was[k]) {
            if (k > second[i] - 'a') ok = 1;
          }
        }
        if (ok) {
          ans += second.substr(i, j - i);
        }
      }
    }
    i = j - 1;
  }
  cnt = 1;
  for (int i = ans.size() - 1; i > 0; --i) {
    if (ans[i] == ans[i - 1])
      cnt++;
    else
      break;
  }
  while (ans.size() > 1 && cnt > was[ans[ans.size() - 1] - 'a'] &&
         ans[ans.size() - 1] == ans[ans.size() - 2]) {
    ans.erase(ans.size() - 1);
    cnt--;
  }
  cout << ans << "\n";
}
