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 = 2e5 + 10;
char s[maxn], ans[maxn];
int n, c[maxn], pref[maxn];
void add(int x) {
  for (; x <= n; x += (x & (-x))) c[x]++;
}
int get(int x) {
  int tot = 0;
  for (; x; x -= (x & (-x))) tot += c[x];
  return tot;
}
char tree[maxn << 2];
void build(int p, int l, int r) {
  if (l >= r) {
    tree[p] = s[l - 1];
    return;
  }
  int m = (r + l) / 2;
  build((p << 1), l, m);
  build(((p << 1) | 1), m + 1, r);
  tree[p] = min(tree[(p << 1)], tree[((p << 1) | 1)]);
}
char query(int p, int l, int r, int s, int t) {
  if (s <= l && r <= t) return tree[p];
  int m = (r + l) / 2;
  char ans = 'z';
  if (s <= m) ans = min(ans, query((p << 1), l, m, s, t));
  if (t > m) ans = min(ans, query(((p << 1) | 1), m + 1, r, s, t));
  return ans;
}
int main() {
  int m;
  scanf("%d", &m);
  if (m < 1 || m > 100000) {
    printf("%d\n", 1 / 0);
    while (true)
      ;
    return 0;
  }
  scanf("%s", s);
  n = strlen(s);
  if (n < 0) {
    printf("%d\n", 1 / 0);
    while (true)
      ;
    return 0;
  }
  build(1, 1, n);
  char rp = 'a';
  for (int i = m; i <= n; i++) rp = max(rp, query(1, 1, n, i - m + 1, i));
  int tot = 0;
  int tmp = 0;
  for (int i = 0; s[i] != 0; i++) {
    if (s[i] < rp) {
      ans[tot++] = s[i];
      add(i + 1);
    }
    if (s[i] == rp) tmp = i;
    pref[i + 1] = tmp + 1;
  }
  int num = 0;
  for (int i = m; i <= n; i++) {
    if (get(i) - get(i - m) == 0) {
      num++;
      add(pref[i]);
    }
  }
  while (num--) ans[tot++] = rp;
  sort(ans, ans + tot);
  ans[tot] = 0;
  printf("%s\n", ans);
  return 0;
}
