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 infl = 4e18;
const long long infi = 1e9;
const long long mod = 1000000007;
auto clk = clock();
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long rng(long long lim) {
  uniform_int_distribution<long long> uid(0, lim - 1);
  return uid(rang);
}
long long powm(long long a, long long b) {
  long long res = 1;
  a %= mod;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n, m;
  string s;
  cin >> m >> s;
  n = ((long long)s.size());
  long long lo = 'a', hi = 'z', mid = (lo + hi) / 2;
  while (lo < hi) {
    bool valid = 1;
    long long last = -1;
    for (long long i = 0; i < n; i++) {
      if (s[i] <= mid) {
        last = i;
      }
      if (i - last >= m) valid = 0;
    }
    if (valid)
      hi = mid;
    else
      lo = mid + 1;
    mid = (lo + hi) / 2;
  }
  long long he = mid, last = -1, lastmota = -1;
  vector<char> ans;
  for (long long i = 0; i < n; i++) {
    if (s[i] < he) {
      last = i;
      ans.push_back(s[i]);
    } else if (s[i] == he) {
      lastmota = i;
    }
    if (i - last >= m) {
      last = lastmota;
      ans.push_back(he);
    }
  }
  sort(ans.begin(), ans.end());
  for (char i : ans) cout << i;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  cout << fixed << setprecision(10);
  long long t = 1;
  for (long long i = 1; i <= t; i++) solve();
  return 0;
}
