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 char nl = '\n';
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const long double EPS = 1e-10;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const int M = 26;
int tot[M];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = s.size();
  for (char c : s) {
    tot[c - 'a']++;
  }
  int last = -1;
  for (int c = 0; c < M; c++) {
    int cur = -1;
    for (int i = 0; i < n; i++) {
      if (s[i] - 'a' <= c) {
        if (i - cur > m) {
          break;
        } else {
          cur = i;
        }
      }
    }
    if (n - cur > m) {
      continue;
    } else {
      last = c;
      break;
    }
  }
  assert(last != -1);
  vector<int> idc, idx;
  for (int i = 0; i < n; i++) {
    if (s[i] - 'a' == last) {
      idc.push_back(i);
    } else if (s[i] - 'a' < last) {
      idx.push_back(i);
    }
  }
  idx.push_back(n);
  function<int(int, int)> solve = [&](int l, int r) {
    int res = 0;
    int rb = r;
    while (rb - l > m) {
      rb = *lower_bound(idc.begin(), idc.end(), rb - m);
      res++;
    }
    return res;
  };
  int pre = -1;
  int cnt = 0;
  for (int i : idx) {
    cnt += solve(pre, i);
    pre = i;
  }
  for (int i = 0; i < last; i++) {
    while (tot[i]--) {
      cout << (char)(i + 'a');
    }
  }
  while (cnt--) {
    cout << (char)(last + 'a');
  }
  cout << nl;
  return 0;
}
