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;
template <class T>
inline T bigmod(T p, T e, T M) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T lcm(T a, T b) {
  a = abs(a);
  b = abs(b);
  return (a / gcd(a, b)) * b;
}
template <class T, class X>
inline bool getbit(T a, X i) {
  T t = 1;
  return ((a & (t << i)) > 0);
}
template <class T, class X>
inline T setbit(T a, X i) {
  T t = 1;
  return (a | (t << i));
}
template <class T, class X>
inline T resetbit(T a, X i) {
  T t = 1;
  return (a & (~(t << i)));
}
inline long long power(long long a, long long b) {
  long long multiply = 1;
  for (int i = (1); i <= (b); i++) {
    multiply *= a;
  }
  return multiply;
}
inline long long ABS(long long a) { return (a >= 0) ? a : -a; }
int m;
string s, s1;
int startpos[26], taken[100003];
bool possible(int pos) {
  int cnt = 0;
  vector<int> lastpos;
  lastpos.push_back(-1);
  int now = s1[pos] - 'a';
  int nowcnt = pos - startpos[now] + 1, lim = 0;
  memset(taken, 0, sizeof(taken));
  for (int i = (0); i <= (m - 1); i++) {
    if (s[i] - 'a' < now) {
      cnt++;
      taken[i] = 1;
    }
    if (s[i] - 'a' == now) lastpos.push_back(i);
  }
  if (cnt == 0 && lastpos[lastpos.size() - 1] < 0) return 0;
  if (cnt == 0) {
    cnt++;
    taken[lastpos[lastpos.size() - 1]] = 1;
    lim++;
  }
  for (int i = (m); i <= (s1.size() - 1); i++) {
    if (s[i] - 'a' < now) {
      cnt++;
      taken[i] = 1;
    }
    if (s[i] - 'a' == now) lastpos.push_back(i);
    if (taken[i - m]) cnt--;
    if (cnt == 0 && lastpos[lastpos.size() - 1] <= i - m) return 0;
    if (cnt == 0) {
      cnt++;
      taken[lastpos[lastpos.size() - 1]] = 1;
      lim++;
    }
  }
  if (lim > nowcnt) return 0;
  return 1;
}
int main() {
  scanf("%d", &m);
  char ch;
  getchar();
  ch = getchar();
  while (ch != '\n') {
    s.push_back(ch);
    ch = getchar();
  }
  s1 = s;
  sort(s1.begin(), s1.end());
  memset(startpos, -1, sizeof(startpos));
  for (int i = (0); i <= (s1.size() - 1); i++) {
    if (startpos[s1[i] - 'a'] == -1) startpos[s1[i] - 'a'] = i;
  }
  int lo = 0, hi = s.size() - 1, mid, best;
  while (hi >= lo) {
    mid = (hi + lo) / 2;
    if (possible(mid)) {
      best = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  for (int i = (0); i <= (best); i++) putchar(s1[i]);
  putchar('\n');
  return 0;
}
