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;
inline double Time() { return (clock() * 1.0) / CLOCKS_PER_SEC; }
const int N = 100500, inf = 1e9 * 2;
const long long MOD = 1e9 + 7ll, INF = 1e18;
const int dx[] = {1, -1, 0, 0, -1, 1, -1, 1};
const int dy[] = {0, 0, 1, -1, -1, 1, 1, -1};
int m, l[N], r[N];
string s;
vector<int> g[26];
bool check(int p) {
  int r1 = 0, l2 = (int((s).size())) - 1;
  if (l[p] >= 0) {
    r1 = min((int((s).size())) - 1, l[p] + m - 1);
  }
  if (r[p] < (int((s).size()))) {
    l2 = max(0, r[p] - m + 1);
  }
  if (l[p] < 0) {
    return l2 == 0;
  }
  if (r[p] > (int((s).size())) - 1) return r1 == (int((s).size())) - 1;
  return (r1 >= r[p] - 1 && l2 <= l[p] + 1);
}
int main() {
  unsigned int FOR;
  asm("rdtsc" : "=A"(FOR));
  srand(FOR);
  cin >> m >> s;
  for (int i = 0; i < (int((s).size())); i++) {
    g[s[i] - 'a'].push_back(i);
  }
  for (int i = 0; i < (int((s).size())); i++) {
    l[i] = i - 1;
    r[i] = i + 1;
  }
  string ans = "";
  bool ok = 0;
  for (int i = 25; i >= 0; i--) {
    bool pp = 0;
    for (auto j : g[i]) {
      if (check(j) && !ok) {
        if (r[j] < (int((s).size()))) l[r[j]] = l[j];
        if (l[j] >= 0) r[l[j]] = r[j];
      } else {
        pp = 1;
        ans += char(i + 'a');
      }
    }
    if (pp) ok = 1;
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
