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 EPS = 1000000007;
const int INF = 1000000000;
double PI = 3.14159265358979323846;
const int MAXN = 1000040;
long long abss(long long h) { return max(h, -h); }
double fabss(double h) { return max(h, -h); }
long long ceill(long long x, long long y) {
  if (x % y != 0) return (x / y) + 1;
  return x / y;
}
string itos(long long num) {
  string str = "";
  if (num == 0) return "0";
  while (num != 0) {
    str += ((num % 10) + '0');
    num /= 10;
  }
  reverse(str.begin(), str.end());
  return str;
}
long long sstoi(string num) {
  long long count = 1;
  for (int i = 0; i < (int)num.size() - 1; i++) {
    count *= 10;
  }
  long long ans = 0;
  for (int i = 0; i < (int)num.size(); i++) {
    ans += (num[i] - '0') * count;
    count /= 10;
  }
  return ans;
}
long long n, m, k, h;
pair<char, int> a[MAXN];
int b[MAXN];
int c[MAXN];
pair<char, int> t[MAXN];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    t[v].first = a[tl].first;
    t[v].second = a[tl].second;
  } else {
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    if (t[v * 2].first > t[v * 2 + 1].first) {
      t[v] = t[v * 2 + 1];
    } else if (t[v * 2].first < t[v * 2 + 1].first) {
      t[v] = t[v * 2];
    } else if (t[v * 2].first == t[v * 2 + 1].first &&
               t[v * 2].second > t[v * 2 + 1].second) {
      t[v] = t[v * 2];
    } else {
      t[v] = t[v * 2 + 1];
    }
  }
}
pair<char, int> sum(int v, int tl, int tr, int l, int r) {
  if (l > r) return make_pair('z', -1);
  if (l == tl && r == tr) return t[v];
  int tm = (tl + tr) / 2;
  pair<char, int> q1 = sum(v * 2, tl, tm, l, min(r, tm));
  pair<char, int> q2 = sum(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  if (q1.first > q2.first) {
    return q2;
  } else if (q1.first < q2.first) {
    return q1;
  } else if (q1.first == q2.first && q1.second > q2.second) {
    return q1;
  } else {
    return q2;
  }
}
int counter[30];
int main() {
  int m;
  cin >> m;
  string str;
  cin >> str;
  int n = str.size();
  for (int i = 0; i < n; i++) {
    a[i].first = str[i];
    a[i].second = i;
    counter[str[i] - 'a']++;
  }
  build(1, 0, n - 1);
  int index = 0;
  vector<char> answer;
  char maxChar = 'a';
  while (true) {
    pair<char, int> w = sum(1, 0, n - 1, index, index + m - 1);
    answer.push_back(w.first);
    maxChar = max(maxChar, w.first);
    index = w.second;
    if (index + m - 1 >= n - 1) {
      break;
    }
    index++;
  }
  sort(answer.begin(), answer.end());
  for (char i = 'a'; i <= 'z'; i++) {
    if (i != maxChar) {
      for (int j = 0; j < counter[i - 'a']; j++) {
        cout << i;
      }
    } else {
      for (int j = 0; j < answer.size(); j++) {
        if (answer[j] == maxChar) {
          cout << maxChar;
        }
      }
      break;
    }
  }
}
