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 maxn = 100010;
int n, m;
char s[maxn];
bool used[maxn];
struct node {
  char val;
  int ind;
  friend bool operator<(node n1, node n2) {
    if (n1.val == n2.val) return n1.ind > n2.ind;
    return n1.val < n2.val;
  }
} tour[4 * maxn];
void build(int i, int l, int r) {
  if (l == r) {
    tour[i] = {s[l], l};
    return;
  }
  int m = (l + r) / 2;
  build(i * 2, l, m);
  build(i * 2 + 1, m + 1, r);
  tour[i] = min(tour[i * 2], tour[i * 2 + 1]);
}
int ql, qr;
node query(int i, int l, int r) {
  if (l >= ql && r <= qr) {
    return tour[i];
  }
  if (l > qr || r < ql) {
    return {'z' + 1, 0};
  }
  int m = (l + r) / 2;
  return min(query(i * 2, l, m), query(i * 2 + 1, m + 1, r));
}
node query(int l, int r) {
  ql = max(0, l), qr = min(n - 1, r);
  return query(1, 0, n - 1);
}
string s2;
int main() {
  scanf("%d %[^\n]", &m, s);
  n = strlen(s);
  build(1, 0, n - 1);
  char mx = 'a';
  int i = 0;
  while (i < n) {
    node q = query(i, i + m - 1);
    s2.push_back(s[q.ind]);
    mx = max(mx, s[q.ind]);
    used[q.ind] = true;
    i = q.ind + 1;
    if (n - i < m) break;
  }
  for (int i = 0; i < n; i++) {
    if (!used[i] && s[i] < mx) s2.push_back(s[i]);
  }
  sort(s2.begin(), s2.end());
  for (char c : s2) {
    printf("%c", c);
  }
  printf("\n");
  return 0;
}
