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 N = 2e5 + 121, T = 4 * N;
int b[N], m, n, i, pos, R, mx;
char a[N], c[N];
string second, ans;
pair<int, int> t[T], P;
void build(int v, int tl, int tr) {
  if (tl == tr)
    t[v] = {b[tl], tl};
  else {
    int tm = (tl + tr) >> 1, one = v << 1, two = one + 1;
    build(one, tl, tm);
    build(two, tm + 1, tr);
    if (t[two].first <= t[one].first)
      t[v] = t[two];
    else
      t[v] = t[one];
  }
}
pair<int, int> get(int v, int tl, int tr, int l, int r) {
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) >> 1, one = v << 1, two = one + 1;
  if (r <= tm) return get(one, tl, tm, l, r);
  if (l > tm) return get(two, tm + 1, tr, l, r);
  pair<int, int> ans1 = get(one, tl, tm, l, r),
                 ans2 = get(two, tm + 1, tr, l, r);
  if (ans2.first <= ans1.first) return ans2;
  return ans1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> m;
  cin >> second;
  n = (int)second.size();
  for (i = 0; i < n; ++i) a[i + 1] = second[i], b[i + 1] = second[i] - 'a';
  build(1, 1, n);
  int last = 1;
  pos = 1;
  while (pos <= n - m + 1) {
    R = pos + m - 1;
    P = get(1, 1, n, pos, R);
    mx = max(mx, P.first);
    c[P.second] = 1;
    pos = P.second + 1;
  }
  ans = "";
  for (i = 1; i <= n; ++i)
    if (b[i] < mx || c[i]) ans += a[i];
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
