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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const int MAXN = 120 * 1000;
int a[MAXN];
int last[MAXN][26];
int num[26];
int main() {
  int m;
  scanf("%d", &m);
  string s = "";
  while (s == "") {
    getline(cin, s);
  }
  int n = ((int)(s).size());
  for (int i = 0; i < n; ++i) {
    a[i] = s[i] - 'a';
  }
  for (int i = 0; i < 26; ++i) {
    last[0][i] = -inf;
  }
  last[0][a[0]] = 0;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < 26; ++j) {
      last[i][j] = last[i - 1][j];
    }
    last[i][a[i]] = i;
  }
  for (int i = 0; i < 26; ++i) {
    num[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    num[a[i]]++;
  }
  for (int i = 0; i < 26; ++i) {
    int cur = 0;
    int numThis = 0;
    bool key = true;
    for (int j = 0; j < n; ++j) {
      cur++;
      if (a[j] < i) {
        cur = 0;
      }
      if (cur == m) {
        cur = j - last[j][i];
        numThis++;
        if (cur >= m) {
          key = false;
          break;
        }
      }
    }
    if (key) {
      for (int j = 0; j < i; ++j) {
        for (int k = 0; k < num[j]; ++k) {
          printf("%c", 'a' + j);
        }
      }
      for (int j = 0; j < numThis; ++j) {
        printf("%c", 'a' + i);
      }
      printf("\n");
      return 0;
    }
  }
  return 0;
}
