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;
int ans[100100], pre[100100], suc[100100];
char c[100100];
int a[100100], n, m;
int main() {
  scanf("%d", &m);
  scanf("%s", c + 1);
  n = strlen(c + 1);
  for (int i = 1; i <= n; i++) a[i] = c[i] - 'a' + 1;
  set<int> s1, s2;
  s1.insert(-100000000);
  s1.insert(100000000);
  s2.insert(-100000000);
  s2.insert(100000000);
  vector<int> g;
  for (int k = 1; k <= 26; k++) {
    g.clear();
    for (int i = 1; i <= n; i++) {
      if (a[i] == k) {
        s1.insert(i);
        g.push_back(i);
      }
    }
    bool ok = true;
    for (int i = 1; i <= n - m + 1; i++) {
      int pos = *s1.lower_bound(i);
      if (pos - i + 1 > m) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      ans[k] = g.size();
      for (int i = 0, sz = g.size(); i < sz; i++) s2.insert(g[i]);
      continue;
    } else {
      vector<int> cur1, cur2;
      cur1.push_back(0);
      for (int i = 1; i <= n; i++)
        if (a[i] == k) cur1.push_back(i);
      cur1.push_back(n + 1);
      for (int i = cur1.size() - 1; i >= 0; i--) cur2.push_back(-cur1[i]);
      for (int i = 1; i <= n; i++) {
        pre[i] = lower_bound(cur2.begin(), cur2.end(), -i) - cur2.begin();
        pre[i] = -cur2[pre[i]];
        suc[i] = lower_bound(cur1.begin(), cur1.end(), i) - cur1.begin();
        suc[i] = cur1[suc[i]];
      }
      for (int i = n - m + 1; i >= 1; i--) {
        int pos = *s2.lower_bound(i);
        if (pos - i + 1 > m) {
          s2.insert(suc[i]);
          ans[k]++;
        }
      }
      break;
    }
  }
  for (int i = 1; i <= 26; i++) {
    for (int j = 0; j < ans[i]; j++) putchar('a' + i - 1);
  }
  putchar('\n');
  return 0;
}
