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 = 1e5 + 42;
int mn[4 * maxn];
const int inf = 1e9;
void upd(int p, int c, int v = 1, int l = 0, int r = maxn) {
  if (r - l == 1) {
    mn[v] = c;
    return;
  }
  int m = (l + r) / 2;
  if (p < m)
    upd(p, c, 2 * v, l, m);
  else
    upd(p, c, 2 * v + 1, m, r);
  mn[v] = min(mn[2 * v], mn[2 * v + 1]);
}
int get(int a, int b, int v = 1, int l = 0, int r = maxn) {
  if (a <= l && r <= b) return mn[v];
  if (r <= a || b <= l) return inf;
  int m = (l + r) / 2;
  return min(get(a, b, 2 * v, l, m), get(a, b, 2 * v + 1, m, r));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  fill(mn, mn + 4 * maxn, inf);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = s.size();
  int sigma = 26;
  int cnt[sigma];
  for (int c = 'a'; c <= 'z'; c++) {
    int dp[n + 1][2];
    dp[0][0] = dp[0][1] = 0;
    upd(0, 0);
    for (int i = 1; i <= n; i++) {
      dp[i][0] = get(max(0, i - m + 1), i);
      if (s[i - 1] == c)
        dp[i][1] = get(max(0, i - m), i) + 1;
      else if (s[i - 1] < c)
        dp[i][1] = get(max(0, i - m), i);
      else
        dp[i][1] = inf;
      upd(i, dp[i][1]);
    }
    if (min(dp[n][0], dp[n][1]) == inf) {
      cout << string(count(begin(s), end(s), c), c);
    } else {
      cout << string(min(dp[n][0], dp[n][1]), c);
      return 0;
    }
  }
  return 0;
}
