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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int m;
char s[N];
struct SparseTable {
  static const int LOG = 20;
  int n;
  pair<char, int> f[LOG][N];
  void init(int _n, pair<char, int>* a) {
    n = _n;
    for (int i = 1; i <= n; ++i) f[0][i] = a[i];
    for (int i = 1; i < LOG; ++i)
      for (int j = 1; j + (1 << i) - 1 <= n; ++j)
        f[i][j] = min(f[i - 1][j], f[i - 1][j + (1 << i - 1)]);
  }
  pair<int, int> RMQ(int l, int r) {
    int k = 31 - __builtin_clz(r - l + 1);
    return min(f[k][l], f[k][r - (1 << k) + 1]);
  }
} st;
pair<char, int> a[N];
int main() {
  ios_base::sync_with_stdio(0);
  while (scanf("%d%s", &m, s + 1) == 2) {
    int n = strlen(s + 1);
    for (int i = 1; i <= n; ++i) {
      a[i] = {s[i], -i};
    }
    st.init(n, a);
    string ans;
    int l = 1, r = m;
    char maxC = 'a';
    vector<int> vis(n + 1, 0);
    while (true) {
      char c;
      int p;
      tie(c, p) = st.RMQ(l, r);
      p = -p;
      vis[p] = true;
      maxC = max(maxC, c);
      ans += c;
      if (p + m - 1 >= n) break;
      l = p + 1;
      r = min(l + m - 1, n);
    }
    for (int i = 1; i <= n; ++i)
      if (!vis[i] && s[i] < maxC) ans += s[i];
    sort(ans.begin(), ans.end());
    cout << ans << endl;
  }
  return 0;
}
