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;
long long mod = 9223372036854775807;
long long fastPow(long long b, long long e) {
  long long r = 1;
  while (e) {
    if (e % 2 == 1) {
      r *= b;
      r %= mod;
    }
    b *= b;
    b %= mod;
    e /= 2;
  }
  return r;
}
long long pgcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else
    return pgcd(b, a % b);
}
long long sign(long long a) {
  if (a < 0) {
    return -1;
  }
  if (a == 0) {
    return 0;
  }
  return 1;
}
bool isPrime(long long a) {
  if (a == 1) {
    return false;
  }
  long long f = sqrt(a);
  for (long long i = 2; i <= f; i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
long long toInt(string s) {
  long long tot = 0;
  for (long long i = s.size() - 1; i >= 0; i--) {
    tot += ((s[i] - '0') % mod) * fastPow(10, i);
    tot %= mod;
  }
  return tot;
}
string toString(long long a) {
  string s = "";
  while (a) {
    s = (char)('0' + a % 10) + s;
    a /= 10;
  }
  return s;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long m;
  cin >> m;
  string s;
  cin >> s;
  set<long long> used;
  vector<char> ans;
  for (long long i = 0; i < s.size(); i++) {
    char c = 'z';
    long long a = -9223372036854775807;
    for (long long j = i; j < i + m; j++) {
      if (c >= s[j]) {
        c = s[j];
        a = j;
      }
    }
    if (used.count(a)) break;
    ans.push_back(c);
    used.insert(a);
    if (a + m >= s.size()) break;
    i = a;
  }
  sort(ans.begin(), ans.end());
  char last = ans.back();
  for (long long i = 0; i < s.size(); i++) {
    if (used.count(i)) continue;
    if (s[i] < last) ans.push_back(s[i]);
  }
  sort(ans.begin(), ans.end());
  for (auto a : ans) {
    cout << a;
  }
  cout << endl;
}
