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 M = 998244353;
const int N = 5 * 1e5 + 5;
const int K = 3000;
const long long INF = 1e18;
const double EPS = 1e-6;
long long k, cnt, u[N];
string second, w;
vector<int> v;
char a;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> k;
  cin >> second;
  string w = "";
  for (int j = 0; j < 26; j++) {
    a = j + 'a';
    cnt = 0;
    for (int i = 0; i < second.size(); i++) {
      if (second[i] == a) {
        u[i] = 1, cnt++;
        v.push_back(i);
      }
    }
    bool ok = 1;
    int l = -1;
    for (int i = 0; i < k; i++) {
      if (u[i]) l = i;
    }
    for (int i = k; i < second.size(); i++) {
      if (l == -1) {
        ok = 0;
        break;
      }
      if (l == i - k && u[i] == 0) {
        ok = 0;
        break;
      }
      if (u[i]) l = i;
    }
    if (ok) {
      break;
    }
  }
  v.push_back(-1);
  v.push_back(second.size());
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size() - 1; i++) {
    if (v[i + 1] - v[i - 1] <= k && second[v[i]] == a) {
      v.erase(v.begin() + i);
      i--;
    }
  }
  for (int i = 1; i < v.size() - 1; i++) w += second[v[i]];
  sort(w.begin(), w.end());
  cout << w;
  return 0;
}
