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 gcd(long long a, long long b) {
  if ((a == 0) || (b == 0)) return a + b;
  return gcd(b, a % b);
}
long long pow_mod(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long pow_2(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long a) {
  for (int i = 3; (i * i) <= a; i += 2) {
    if ((a % i) == 0) return false;
  }
  if ((a != 2) && ((a % 2) == 0)) return false;
  return true;
}
string con_ll_to_str(int a) {
  stringstream mystr;
  mystr << a;
  return mystr.str();
}
long long con_str_to_ll(string st) {
  long long numb = 0;
  int len = st.size(), i, j = 0;
  for (i = len - 1; i >= 0; i--) {
    numb += (pow_2(10, j) * (st[i] - '0'));
    j++;
  }
  return numb;
}
long long baseno_to_decimal(string st, long long basee) {
  long long i = 0, j, anss = 0;
  for (j = (int)st.length() - 1; j >= 0; j--) {
    anss += (st[j] - '0') * pow_2(basee, i);
    i++;
  }
  return anss;
}
string decimal_to_string(long long num, long long basee) {
  long long i = 0, j, opop;
  string anss = "";
  vector<string> stri;
  stri.push_back("0");
  stri.push_back("1");
  stri.push_back("2");
  stri.push_back("3");
  stri.push_back("4");
  stri.push_back("5");
  stri.push_back("6");
  stri.push_back("7");
  stri.push_back("8");
  stri.push_back("9");
  stri.push_back("A");
  stri.push_back("B");
  stri.push_back("C");
  stri.push_back("D");
  stri.push_back("E");
  stri.push_back("F");
  while (num) {
    opop = num % basee;
    anss += stri[opop];
    num /= basee;
  }
  reverse(anss.begin(), anss.end());
  return anss;
}
long long dp[26];
bool flag;
vector<int> vc[26], vc2;
set<char> se;
long long check(char ch, long long pre, long long M) {
  long long i, j, pos = pre;
  long long star = lower_bound(vc[ch - 'a'].begin(), vc[ch - 'a'].end(), pre) -
                   vc[ch - 'a'].begin();
  for (i = star; i <= (int)vc[ch - 'a'].size() - 1; i++) {
    if ((pre + M - 1) >= vc[ch - 'a'][i]) {
      flag = 1;
      pos = vc[ch - 'a'][i];
    } else {
      break;
    }
  }
  return pos;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long T, N, M, i, j, opop;
  string st, st2;
  cin >> M;
  cin >> st;
  N = st.length();
  for (i = 0; i <= N - 1; i++) {
    vc[st[i] - 'a'].push_back(i);
  }
  long long pos = 0, prev = -1;
  while (1) {
    for (j = 0; j <= 25; j++) {
      flag = 0;
      pos = check(j + 'a', prev + 1, M);
      if (flag) {
        prev = pos;
        se.insert(st[pos]);
        st2.push_back(st[pos]);
        break;
      }
    }
    if ((prev + M) >= N) break;
  }
  for (auto it = se.begin(); it != se.end(); ++it) {
    opop = *it - 'a';
  }
  for (i = 0; i <= (int)st2.size() - 1; i++) {
    dp[st2[i] - 'a']++;
  }
  for (i = 0; i < opop; i++) {
    for (j = 0; j < (vc[i].size() - dp[i]); j++) {
      st2.push_back('a' + i);
    }
  }
  sort(st2.begin(), st2.end());
  cout << st2;
  return 0;
}
