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 inf = 1234567890;
const long long infl = 8876543210123456789ll;
const int mod = 1000000007;
void init() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(17);
  cout << fixed;
}
template <class RandomAccessIterator>
void print(RandomAccessIterator a, RandomAccessIterator b) {
  for (; a < b; ++a) {
    cout << (*a) << '\n';
  }
}
template <class RandomAccessIterator>
void println(RandomAccessIterator a, RandomAccessIterator b) {
  for (; a < b; ++a) {
    cout << (*a) << ' ';
  }
  cout << '\n';
}
template <class RandomAccessIterator>
void rarr(RandomAccessIterator a, RandomAccessIterator b) {
  for (; a < b; ++a) {
    cin >> (*a);
  }
}
template <class RandomAccessIterator>
void rgraph(RandomAccessIterator a, int vert, int edj) {
  for (int i = 0; i < (edj); ++i) {
    int u, v;
    cin >> u >> v;
    a[--u].push_back(--v);
    a[v].push_back(u);
  }
}
template <class RandomAccessIterator>
void rdgraph(RandomAccessIterator a, int vert, int edj) {
  for (int i = 0; i < (edj); ++i) {
    int u, v;
    cin >> u >> v;
    a[--u].push_back(--v);
  }
}
template <typename T>
T vmin(T&& t) {
  return std::forward<T>(t);
}
template <typename T0, typename T1, typename... Ts>
typename std::common_type<T0, T1, Ts...>::type vmin(T0&& val1, T1&& val2,
                                                    Ts&&... vs) {
  if (val2 < val1)
    return vmin(val2, std::forward<Ts>(vs)...);
  else
    return vmin(val1, std::forward<Ts>(vs)...);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int m;
string s;
vector<int> p[26];
int ns[26][100001];
int f(char c) { return c - 'a'; }
int main() {
  init();
  cin >> m >> s;
  for (int i = 0; i < (s.size()); ++i) p[f(s[i])].push_back(i);
  for (int a = 0; a < (26); ++a) {
    char c = 'a' + a;
    for (int i = 0; i < (m); ++i) {
      if (s[i] < c) ++ns[a][0];
    }
    for (int i = (1); i < (s.size() + 1 - m); ++i) {
      ns[a][i] = ns[a][i - 1];
      if (s[i - 1] < c) --ns[a][i];
      if (s[i + m - 1] < c) ++ns[a][i];
    }
  }
  for (char c = 'a'; c <= 'z'; c++) {
    if (p[f(c)].size() > 0) {
      int a = f(c);
      int r = 0;
      int j = 0;
      int lc = -1;
      bool fail = false;
      for (int i = 0; i < (s.size() + 1 - m); ++i) {
        if (ns[a][i] == 0 && lc < i) {
          while (j < p[a].size() && p[a][j] < i + m) {
            ++j;
          }
          --j;
          if (j < 0 || p[a][j] < i) {
            fail = true;
            break;
          } else {
            lc = p[a][j];
            ++r;
          }
        }
      }
      if (!fail) {
        string ans = "";
        for (char d = 'a'; d < c; ++d) {
          for (int v = 0; v < (p[f(d)].size()); ++v) ans.push_back(d);
        }
        for (int v = 0; v < (r); ++v) ans.push_back(c);
        cout << (ans) << '\n';
        return 0;
      }
    }
  }
}
