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 MAXN = 200005;
char s[MAXN];
int m, n;
vector<int> ve[60];
set<pair<int, int> > st;
char ans[MAXN];
int sz = 0;
int main() {
  cin >> m;
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) ve[s[i] - 'a'].push_back(i);
  st.insert({0, n - m});
  for (int i = 0; i < 30; i++) {
    set<pair<int, int> >::iterator it = st.begin();
    int l = 0, r = 0, cnt = 0;
    for (int j = 0; j < ve[i].size(); j++) {
      if (it == st.end()) break;
      l = it->first, r = it->second;
      while (r + m <= ve[i][j]) {
        ++it;
        if (it == st.end()) break;
        l = it->first, r = it->second;
      }
      if ((r + m) <= ve[i][j]) break;
      if (ve[i][j] < l) continue;
      if (j + 1 < ve[i].size() && ve[i][j + 1] <= l + m - 1)
        continue;
      else {
        cnt++;
        st.erase(it);
        if (ve[i][j] - m >= l) st.insert({l, ve[i][j] - m});
        if (ve[i][j] + 1 <= r) st.insert({ve[i][j] + 1, r});
        it = st.lower_bound({ve[i][j] + 1, -1});
      }
    }
    if (st.size() > 0) cnt = ve[i].size();
    for (int j = 0; j < cnt; j++) ans[sz++] = 'a' + i;
    if (st.size() == 0) break;
  }
  printf("%s\n", ans);
  return 0;
}
