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 Dim = 8e5 + 5;
pair<int, int> tree[Dim];
int n, m, c[Dim];
char s[Dim];
vector<char> ans;
void prec(int inc, int sf, int node) {
  if (inc == sf) {
    tree[node] = {s[inc] - 'a' + 1, -inc};
    return;
  }
  int mid = (inc + sf) / 2;
  prec(inc, mid, 2 * node);
  prec(mid + 1, sf, 2 * node + 1);
  tree[node] = min(tree[2 * node], tree[2 * node + 1]);
}
pair<int, int> query(int inc, int sf, int a, int b, int node) {
  if (a <= inc && sf <= b) {
    return tree[node];
  }
  pair<int, int> ans = {Dim, -1};
  int mid = (inc + sf) / 2;
  if (a <= mid) {
    ans = min(ans, query(inc, mid, a, b, 2 * node));
  }
  if (b > mid) {
    ans = min(ans, query(mid + 1, sf, a, b, 2 * node + 1));
  }
  return ans;
}
int main() {
  cin >> m >> (s + 1);
  n = strlen(s + 1);
  prec(1, n, 1);
  int lst = 0;
  for (int i = 1; i <= n - m + 1; ++i) {
    int pos = -query(1, n, i, i + m - 1, 1).second;
    if (s[pos] == s[lst] && lst >= i) {
      continue;
    }
    lst = pos;
    if (!c[pos]) {
      c[pos] = 1;
      ans.push_back(s[pos]);
    }
  }
  sort(ans.begin(), ans.end());
  char cmax = ans[ans.size() - 1];
  for (int i = 1; i <= n; ++i) {
    if (s[i] < cmax && c[i] == 0) {
      ans.push_back(s[i]);
    }
  }
  sort(ans.begin(), ans.end());
  for (vector<char>::iterator it = ans.begin(); it != ans.end(); ++it) {
    cout << *it;
  }
  return 0;
}
