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>
const int maxn = 2e5 + 10;
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qpow(long long p, long long q) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p;
    p = p * p;
    q >>= 1;
  }
  return f;
}
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, k, t, vis[maxn];
struct node {
  char x;
  int pos;
  node(char _x, int _pos) : x(_x), pos(_pos) {}
  bool operator<(const node& p) const {
    return x == p.x ? pos > p.pos : x < p.x;
  }
};
char a[maxn], ma;
string ans;
set<node> pq;
int main() {
  int i, j;
  scanf("%d%s", &n, a);
  int len = strlen(a), pre = 0;
  for (i = 0; i < n; i++) pq.insert(node(a[i], i));
  i = n;
  while (1) {
    node now = *pq.begin();
    ans += now.x;
    ma = max(ma, now.x);
    vis[now.pos] = 1;
    for (j = pre; j <= now.pos; j++) pq.erase(node(a[j], j));
    pre = now.pos + 1;
    int cnt = 0;
    for (j = i; j < len && j - now.pos <= n; j++) {
      i = j;
      cnt = j - now.pos;
      pq.insert(node(a[j], j));
    }
    i++;
    if (cnt < n) break;
  }
  for (i = 0; a[i]; i++)
    if (!vis[i] && a[i] < ma) ans += a[i];
  sort(ans.begin(), ans.end());
  cout << ans << endl;
  return 0;
}
