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 INF = 1 << 30;
const int N = 1e5 + 10;
char s[N];
int pre[N][26];
int main() {
  int m, n;
  scanf("%d%s", &m, s);
  n = strlen(s);
  for (int k = 0; k < 26; ++k) {
    int last = -1;
    for (int i = 0; i < n; ++i) {
      if (s[i] - 'a' == k) last = i;
      pre[i][k] = last;
    }
    pre[n][k] = last;
  }
  vector<int> A = {-1, n};
  char c;
  vector<char> ret;
  for (c = 'a'; c <= 'z'; ++c) {
    int cnt = 0;
    vector<int> B = A;
    for (int i = 0; i < n; ++i) {
      if (s[i] == c) {
        B.push_back(i);
        ++cnt;
      }
    }
    sort(B.begin(), B.end());
    bool found = false;
    for (int i = 1; i < B.size(); ++i) {
      if (B[i] - B[i - 1] > m) {
        found = true;
        break;
      }
    }
    if (!found) break;
    for (int i = 0; i < cnt; ++i) ret.push_back(c);
    A = move(B);
  }
  for (int i = 0; i < A.size() - 1; ++i) {
    if (A[i + 1] - A[i] <= m) continue;
    for (int k = A[i]; k + m < A[i + 1];) {
      k = pre[k + m][c - 'a'];
      ret.push_back(c);
    }
  }
  for (auto& it : ret) {
    putchar(it);
  }
  puts("");
  return 0;
}
