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;
struct info {
  long long int val, in;
};
long long int arr[100010];
info tree[100010 * 4], temp;
void init(long long int node, long long int b, long long int e) {
  if (b == e) {
    tree[node].val = arr[b];
    tree[node].in = b;
    return;
  }
  long long int Left = node * 2;
  long long int Right = node * 2 + 1;
  long long int mid = (b + e) / 2;
  init(Left, b, mid);
  init(Right, mid + 1, e);
  if (tree[Left].val >= tree[Right].val) {
    tree[node].val = tree[Right].val;
    tree[node].in = tree[Right].in;
  } else {
    tree[node].val = tree[Left].val;
    tree[node].in = tree[Left].in;
  }
}
info query(long long int node, long long int b, long long int e,
           long long int i, long long int j) {
  if (i > e || j < b) {
    temp.in = temp.val = 30;
    return temp;
  }
  if (b >= i && e <= j) return tree[node];
  long long int Left = node * 2;
  long long int Right = node * 2 + 1;
  long long int mid = (b + e) / 2;
  info p1 = query(Left, b, mid, i, j);
  info p2 = query(Right, mid + 1, e, i, j);
  if (p1.val >= p2.val)
    return p2;
  else
    return p1;
}
int main() {
  long long int i, j, len, mn = 30, m, last = 1, in;
  long long int used = -1;
  string s;
  vector<char> v, ans;
  char mxx = 'a' - 1;
  cin >> m >> s;
  len = s.length();
  for (i = 0; i < s.length(); i++) {
    arr[i + 1] = s[i] - 'a' + 1;
  }
  init(1, 1, len);
  for (i = 1; i <= len - m + 1; i++) {
    info ret = query(1, 1, len, i, i + m - 1);
    if (used == -1) {
      used = ret.in;
      i = ret.in;
      last = ret.val;
      v.push_back(last - 1 + 'a');
      mxx = max(mxx, (char)(last - 1 + 'a'));
    } else {
      if (ret.val == last && used == ret.in) {
      } else {
        used = ret.in;
        i = ret.in;
        last = ret.val;
        v.push_back(last - 1 + 'a');
        mxx = max(mxx, (char)(last - 1 + 'a'));
      }
    }
  }
  if (v.size() == 0) v.push_back(s[0]);
  sort(v.begin(), v.end());
  long long int inx = v.size() - 1;
  long long int cnt = 0;
  for (i = 0; i < v.size(); i++) {
    if (v[i] == v[inx]) cnt++;
  }
  for (i = 0; i < s.length(); i++) {
    if (s[i] < mxx) ans.push_back(s[i]);
  }
  for (i = 1; i <= cnt; i++) {
    ans.push_back(v[inx]);
  }
  sort(ans.begin(), ans.end());
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i];
  }
  cout << endl;
  return 0;
}
