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) { return b ? gcd(b, a % b) : a; }
long long powM(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = powM(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long power(long long x, long long y, long long p = LLONG_MAX) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool cst(const pair<long long, long long>& a,
         const pair<long long, long long>& b) {
  return abs(a.first - a.second) > abs(b.first - b.second);
}
long long mod = 998244353;
long long inv(long long val, long long MODx = mod) {
  return power(val, MODx - 2, MODx);
}
vector<long long> fac, ifac;
void preFac(long long sz) {
  fac.resize(sz + 1), ifac.resize(sz + 1);
  fac[0] = 1;
  for (int i = 1; i <= sz; i++) {
    fac[i] = (i * fac[i - 1]) % mod;
  }
  ifac[sz] = inv(fac[sz]);
  for (int i = sz - 1; i >= 0; i--) {
    ifac[i] = ((i + 1) * ifac[i + 1]) % mod;
  }
}
long long nCr(long long N, long long R) {
  if (R <= N && R >= 0) {
    return ((fac[N] * ifac[R]) % mod * ifac[N - R]) % mod;
  }
  return 0;
}
long long fn(long long x, long long rn[]) {
  if (x == rn[x])
    return x;
  else
    return rn[x] = fn(rn[x], rn);
}
bool un(long long x, long long y, long long rn[], long long sz[]) {
  x = fn(x, rn);
  y = fn(y, rn);
  if (x == y) return false;
  if (sz[x] < sz[y]) swap(x, y);
  sz[x] += sz[y];
  rn[y] = x;
  return true;
}
long long num = 0;
pair<long long, long long> t[4 * 100001];
void build(long long a[], long long v, long long tl, long long tr) {
  if (tl == tr)
    t[v] = {a[tl], 1};
  else {
    long long tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
    if (t[v * 2].first == t[v * 2 + 1].first)
      t[v] = {t[2 * v].first, t[2 * v].second + t[2 * v + 1].second};
    else if (t[v * 2].first < t[v * 2 + 1].first)
      t[v] = {t[2 * v].first, t[2 * v].second};
    else
      t[v] = {t[2 * v + 1].first, t[2 * v + 1].second};
  }
}
pair<long long, long long> query(long long v, long long tl, long long tr,
                                 long long l, long long r) {
  if (tl > r || tr < l) return {1000000007, 1};
  if (tl >= l && tr <= r) return t[v];
  long long tm = (tl + tr) / 2;
  pair<long long, long long> q1 = query(2 * v, tl, tm, l, r);
  pair<long long, long long> q2 = query(2 * v + 1, tm + 1, tr, l, r);
  if (q1.first == q2.first)
    return {q1.first, q1.second + q2.second};
  else if (q1.first < q2.first)
    return {q1.first, q1.second};
  else
    return {q2.first, q2.second};
}
void update(long long v, long long tl, long long tr, long long pos,
            long long new_val) {
  if (tl == tr)
    t[v] = {new_val, 1};
  else {
    long long tm = (tl + tr) / 2;
    if (pos <= tm)
      update(v * 2, tl, tm, pos, new_val);
    else
      update(v * 2 + 1, tm + 1, tr, pos, new_val);
    if (t[v * 2].first == t[v * 2 + 1].first)
      t[v] = {t[2 * v].first, t[2 * v].second + t[2 * v + 1].second};
    else if (t[v * 2].first < t[v * 2 + 1].first)
      t[v] = {t[2 * v].first, t[2 * v].second};
    else
      t[v] = {t[2 * v + 1].first, t[2 * v + 1].second};
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m;
  cin >> m;
  string s;
  cin >> s;
  int n = s.length();
  s = '.' + s;
  map<int, set<int>> mp;
  for (long long i = 1; i <= n; i++) mp[s[i] - 'a'].insert(i);
  int fl = 0;
  set<int> st;
  st.insert(0);
  st.insert(n + 1);
  for (long long j = 0; j <= 25; j++) {
    mp[j].insert(0);
    if ((int)mp[j].size() == 1) continue;
    for (auto u : st) {
      if (u == n + 1) {
        fl = 1;
        break;
      }
      auto it = st.find(u);
      int t = *(++it);
      if (t - u <= m) continue;
      auto f = *(--mp[j].upper_bound(u + m));
      if (f >= u + 1 && f <= u + m) {
        st.insert(f);
      } else {
        break;
      }
    }
    if (fl == 1) break;
    for (auto u : mp[j]) st.insert(u);
  }
  multiset<char> ms;
  for (auto u : st) {
    if (u == 0 || u == n + 1) continue;
    ms.insert(s[u]);
  }
  for (auto u : ms) cout << u;
  return 0;
}
