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 MAX = 200005;
const int MOD = 1000000007;
const int MAGIC = 3;
struct BIT {
  int t[MAX];
  void upd(int i, int v) {
    if (i) {
      for (; i < MAX; i += (i & (-i))) {
        t[i] += v;
      }
    }
  }
  long long get(int i) {
    long long second = 0;
    for (; i > 0; i -= (i & (-i))) {
      second += t[i];
    }
    return second;
  }
};
int a[MAX];
string second;
int m;
int n;
bool ok(int x) {
  vector<int> all;
  all.push_back(-1);
  for (int i = 0; i < n; i++) {
    if (a[i] <= x) {
      all.push_back(i);
    }
  }
  all.push_back(n);
  for (int i = 1; i < all.size(); i++) {
    if (all[i] - all[i - 1] > m) {
      return false;
    }
  }
  return true;
}
bool choose[MAX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m;
  cin >> second;
  n = second.length();
  for (int i = 0; i < n; i++) {
    a[i] = second[i] - 'a';
  }
  int lo = 0;
  int hi = 25;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    if (!ok(mid)) {
      lo = mid + 1;
    } else {
      hi = mid;
    }
  }
  BIT b;
  vector<int> opts;
  for (int i = 0; i < n; i++) {
    if (a[i] < lo) {
      choose[i] = true;
      b.upd(i + 1, 1);
      b.upd(i + m + 1, -1);
    }
    if (a[i] == lo) {
      opts.push_back(i);
    }
  }
  vector<int> need;
  for (int i = m; i < n + 1; i++) {
    if (b.get(i) == 0) {
      need.push_back(i - 1);
    }
  }
  for (int i = need.size() - 1; i >= 0; i--) {
    if (b.get(need[i] + 1) != 0) {
      continue;
    }
    int it =
        lower_bound(opts.begin(), opts.end(), need[i] - m + 1) - opts.begin();
    choose[opts[it]] = true;
    b.upd(opts[it] + 1, 1);
    b.upd(opts[it] + m + 1, -1);
  }
  vector<char> v;
  for (int i = 0; i < n; i++) {
    if (choose[i]) {
      v.push_back(second[i]);
    }
  }
  sort(v.begin(), v.end());
  for (char i : v) {
    cout << i;
  }
}
