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 long long infi = 1000000000000000009;
char str[100009 + 3];
vector<int> v[30];
int c[30];
int h[30];
int t[30];
int an[100009 + 5];
int main() {
  int m, n, i, j, k, l;
  scanf("%d", &m);
  scanf("%s", str);
  n = strlen(str);
  for (i = 0; i < n; i++) v[str[i] - 'a'].push_back(i);
  int curr = 0, maxi = 0;
  for (i = 0; i <= n - m; i++) {
    j = i + m - 1;
    while (curr <= j) {
      t[str[curr] - 'a']++;
      curr++;
    }
    for (k = 0; k < 26; k++) {
      if (t[k] != 0) break;
    }
    maxi = max(maxi, k);
    an[i] = k;
    t[str[i] - 'a']--;
  }
  for (i = 0; i < 26; i++) v[i].push_back(n);
  stack<pair<int, int> > st;
  for (i = 0; i <= n - m; i++) {
    j = i + m - 1;
    k = an[i];
    if (an[i] == maxi) {
      while (v[k][c[k]] != n && v[k][c[k]] < i) c[k]++;
      int ans = v[k][c[k]];
      if (st.empty()) {
        st.push(make_pair(ans, j));
        h[str[ans] - 'a']++;
      } else {
        int a = st.top().first, b = st.top().second;
        if (ans <= b && str[ans] == str[a]) {
          st.pop();
          st.push(make_pair(ans, b));
        } else {
          st.push(make_pair(ans, j));
          h[str[ans] - 'a']++;
        }
      }
    }
  }
  for (i = 0; i < n; i++) {
    if ((str[i] - 'a') < maxi) h[str[i] - 'a']++;
  }
  for (i = 0; i < 26; i++) {
    for (j = 1; j <= h[i]; j++) printf("%c", i + 'a');
  }
  return 0;
}
