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 long long INF = (int)1e9;
const long long LINF = (long long)1e18;
const double EPS = 1e-9;
const int N = 1e5 + 123;
const int L = 'z' - 'a' + 1;
int n, m;
int a[N];
int cnt[L];
vector<int> pos[L];
bool ok[N];
int run() {
  cin >> m;
  string s;
  cin >> s;
  n = (int)s.size();
  for (int i = 1; i <= n; i++) {
    a[i] = s[i - 1] - 'a';
    cnt[a[i]]++;
    pos[a[i]].push_back(i);
  }
  set<int> q;
  for (int i = 0; i <= n + 1; i++) {
    q.insert(i);
  }
  for (int i = L - 1; i >= 0; i--) {
    bool flag = true;
    sort(pos[i].begin(), pos[i].end());
    for (int p : pos[i]) {
      auto it = q.find(p);
      int prv, nxt;
      --it;
      prv = *it;
      ++it;
      ++it;
      nxt = *it;
      if (nxt - prv <= m) {
        q.erase(p);
        cnt[i]--;
        ok[p] = true;
      } else {
        flag = false;
      }
    }
    if (!flag) {
      break;
    }
  }
  for (int i = 0; i < L; i++) {
    for (int j = 0; j < cnt[i]; j++) {
      cout << ((char)(i + 'a'));
    }
  }
  cout << "\n";
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  return run();
}
