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 N = 1e5 + 5;
const int SN = 1 << 18;
int n, m;
char str[N];
int taken[N];
string ans = "";
int main() {
  scanf("%d", &m);
  scanf("%s", str + 1);
  n = strlen(str + 1);
  for (int i = 1; i <= n; ++i) {
    taken[i] = 0;
  }
  for (char c = 'a'; c <= 'z'; ++c) {
    int last = 0;
    int idx = -1;
    vector<int> tmp;
    tmp.clear();
    bool rekt = 0;
    for (int i = 1; i <= n; ++i) {
      if (str[i] == c) {
        idx = i;
      }
      if (taken[i]) {
        last = i;
      }
      if (i - last >= m) {
        last = idx;
        if (i - idx >= m) {
          rekt = 1;
          break;
        }
        tmp.emplace_back(idx);
      }
    }
    if (rekt) {
      for (int i = 1; i <= n; ++i) {
        if (str[i] == c) {
          ans += c;
          taken[i] = 1;
        }
      }
    } else {
      for (auto it : tmp) {
        ans += c;
      }
      break;
    }
  }
  cout << ans << endl;
}
