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;
int N;
string s;
int pre[26][100000];
int no[26];
int32_t main() {
  ios_base::sync_with_stdio(false);
  if (fopen("cf724d.in", "r")) {
    freopen("cf724d.in", "r", stdin);
    freopen("cf724d.out", "w", stdout);
  }
  cin >> N >> s;
  for (char i = 'a'; i <= 'z'; i++) {
    pre[i - 'a'][0] = (bool)(s[0] <= i);
    for (int j = 1; j < s.length(); j++) {
      pre[i - 'a'][j] = pre[i - 'a'][j - 1] + (bool)(s[j] <= i);
    }
  }
  for (int i = 0; i < 26; i++) {
    bool trash = false;
    if (pre[i][N - 1] == 0) {
      no[i] = pre[i][s.length() - 1];
      if (i) {
        no[i] -= pre[i - 1][s.length() - 1];
      }
      continue;
    }
    for (int j = N; j < s.length(); j++) {
      if (pre[i][j] - pre[i][j - N] == 0) {
        no[i] = pre[i][s.length() - 1];
        if (i) {
          no[i] -= pre[i - 1][s.length() - 1];
        }
        trash = true;
        break;
      }
    }
    if (trash) {
      continue;
    }
    int prevtaken = -1, preval;
    for (int j = 0; j < s.length(); j++) {
      if (j - prevtaken > N) {
        prevtaken = preval;
        no[i]++;
      }
      if (s[j] != 'a' + i) {
        if (s[j] < 'a' + i) {
          prevtaken = j;
        }
        continue;
      }
      preval = j;
    }
    if (s.length() - prevtaken > N) {
      no[i]++;
    }
    break;
  }
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < no[i]; j++) {
      cout << (char)('a' + i);
    }
  }
  cout << '\n';
  return 0;
}
