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;
long long m;
string str;
long long ans[30];
set<long long> sendAll(char c) {
  set<long long> ret;
  for (long long j = 0; j < str.size(); ++j) {
    if (str[j] == c) {
      ret.insert(j);
    }
  }
  return ret;
}
set<long long> getIdx(char c) {
  set<long long> idx;
  long long pre = -1;
  long long preselec = -1;
  long long gap = 0;
  long long pregap = 0;
  for (long long i = 0; i < str.size(); ++i) {
    if (str[i] != '*') {
      gap++;
      pregap++;
    } else {
      gap = 0;
      pregap = 0;
    }
    if (str[i] == c) {
      pre = i;
      pregap = 0;
    }
    if (gap >= m) {
      if (pre == preselec) {
        return sendAll(c);
      }
      idx.insert(pre);
      preselec = pre;
      gap = pregap;
    }
  }
  string temp = str;
  for (set<long long>::iterator it = idx.begin(); it != idx.end(); ++it) {
    temp[*it] = '*';
  }
  long long cont = 0;
  for (long long i = 0; i < temp.size(); ++i) {
    if (cont >= m) {
      return sendAll(c);
    }
    if (temp[i] != '*') {
      cont++;
    } else {
      cont = 0;
    }
  }
  return idx;
}
void pruneStr(set<long long> idx) {
  for (set<long long>::iterator it = idx.begin(); it != idx.end(); ++it) {
    str[*it] = '*';
  }
}
int main() {
  cin >> m >> str;
  for (long long i = 0; i < 26; ++i) {
    ans[i] = 0;
  }
  for (long long i = 0; i < 26; ++i) {
    set<long long> idx = getIdx((char)(i + 'a'));
    ans[i] = idx.size();
    pruneStr(idx);
    if (str == "") {
      break;
    }
  }
  for (long long i = 0; i < 26; ++i) {
    while (ans[i]--) {
      cout << (char)('a' + i);
    }
  }
  cout << endl;
}
