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;
bool taken[100005];
long long ct[26];
long long m;
string s;
signed main() {
  long long t = 1;
  ;
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(NULL);
  cout << fixed << setprecision(10);
  while (t-- > 0) {
    memset(ct, 0, sizeof(ct));
    memset(taken, false, sizeof(taken));
    cin >> m >> s;
    long long ct_needed;
    long long n = s.length();
    long long i, p, li = 0, lt;
    for (long long i = 0; i < n; i++) {
      ct[s[i] - 'a']++;
    }
    bool flag1 = false;
    for (i = 0; i < 26; i++) {
      ct_needed = 0;
      li = -1, lt = -1;
      char srch = 'a' + i;
      p = 0;
      flag1 = false;
      for (long long j = 0; j < m && (p + j) < (n); j++) {
        if (taken[p + j]) {
          lt = p + j;
          p = p + j + 1;
          j = -1;
          continue;
        }
        if (s[p + j] == srch) {
          li = p + j;
        }
        if ((p + j - lt) >= m) {
          if ((p + j - li) >= m) {
            flag1 = true;
            break;
          } else {
            taken[li] = true;
            lt = li;
            p = p + j;
            j = -1;
            ct_needed++;
          }
        }
      }
      if (flag1) {
        for (long long j = 0; j < n; j++) {
          if (s[j] == ('a' + i)) {
            taken[j] = true;
          }
        }
      };
      if (!flag1) {
        break;
      }
    }
    string ans = "";
    for (long long r = 0; r < i; r++) {
      for (long long j = 0; j < ct[r]; j++) {
        ans += ('a' + r);
      }
    }
    for (long long j = 0; j < ct_needed; j++) {
      ans += ('a' + i);
    }
    cout << ans << "\n";
  };
  return 0;
}
