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>
struct Seg {
  const T ID = 0;
  T comb(T a, T b) { return a + b; }
  long long n;
  vector<T> seg;
  void init(long long _n) {
    n = _n;
    seg.assign(2 * n, ID);
  }
  void pull(long long p) { seg[p] = comb(seg[2 * p], seg[2 * p + 1]); }
  void upd(long long p, T val) {
    seg[p += n] = val;
    for (p /= 2; p; p /= 2) pull(p);
  }
  T query(long long l, long long r) {
    T ra = ID, rb = ID;
    for (l += n, r += n + 1; l < r; l /= 2, r /= 2) {
      if (l & 1) ra = comb(ra, seg[l++]);
      if (r & 1) rb = comb(seg[--r], rb);
    }
    return comb(ra, rb);
  }
};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long m;
  cin >> m;
  string s;
  cin >> s;
  long long n = s.size();
  long long a[n];
  for (long long i = 0; i < n; ++i) {
    a[i] = s[i] - 'a';
  }
  vector<long long> ans;
  set<array<long long, 2>> ranges;
  Seg<long long> st;
  st.init(n);
  for (long long i = 0; i < n; ++i) {
    if (i + m - 1 >= n) break;
    ranges.insert({i, i + m - 1});
  }
  for (long long i = 0; i < 26; ++i) {
    bool fin = 1;
    for (long long j = 0; j < n; ++j) {
      if (a[j] == i) {
        st.upd(j, 1);
      }
    }
    for (auto [l, r] : ranges) {
      fin &= st.query(l, r) > 0;
    }
    if (fin) {
      vector<long long> pos;
      for (long long j = 0; j < n; ++j) {
        if (a[j] == i) {
          st.upd(j, 0);
          pos.push_back(j);
        }
      }
      for (auto [l, r] : ranges) {
        if (st.query(l, r)) continue;
        auto lb = --(lower_bound((pos).begin(), (pos).end(), r + 1));
        st.upd(*lb, 1);
        ans.push_back(i);
      }
      break;
    } else {
      for (long long j = 0; j < n; ++j) {
        if (a[j] == i) {
          ans.push_back(i);
        }
      }
      vector<array<long long, 2>> rem;
      for (auto [l, r] : ranges) {
        if (st.query(l, r)) {
          rem.push_back({l, r});
        }
      }
      for (auto [l, r] : rem) {
        ranges.erase({l, r});
      }
    }
  }
  for (long long i : ans) {
    cout << (char)(i + 'a');
  }
  cout << "\n";
}
