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 int N = 1e5 + 10, INF = 0x3f3f3f3f;
char s[N];
struct node {
  int pos, x;
} tree[N << 2];
vector<char> v;
int cnt[30];
void pushUp(int t) {
  if (tree[t << 1].x < tree[t << 1 | 1].x) {
    tree[t].x = tree[t << 1].x;
    tree[t].pos = tree[t << 1].pos;
  } else {
    tree[t].x = tree[t << 1 | 1].x;
    tree[t].pos = tree[t << 1 | 1].pos;
  }
}
void update(int t, int L, int R, int x, int v) {
  if (L == R && L == x) {
    tree[t].x = v;
    tree[t].pos = x;
    return;
  }
  int mid = (L + R) >> 1;
  if (x <= mid)
    update(t << 1, L, mid, x, v);
  else
    update(t << 1 | 1, mid + 1, R, x, v);
  pushUp(t);
}
int main() {
  int m, i;
  scanf("%d%s", &m, s);
  memset(tree, 0x3f, sizeof tree);
  int n = strlen(s), pos = -INF, mins = 26;
  for (i = 0; i < min(m - 1, n); ++i) {
    update(1, 1, n, i + 1, s[i] - 'a');
    cnt[s[i] - 'a']++;
  }
  for (i = m - 1; i < n; ++i) {
    cnt[s[i] - 'a']++;
    update(1, 1, n, i + 1, s[i] - 'a');
    if (i - m >= 0) update(1, 1, n, i - m + 1, INF);
    if (pos >= i - m + 1 && pos <= i) continue;
    pos = tree[1].pos - 1;
    v.push_back(s[pos]);
  }
  sort(v.begin(), v.end());
  i = 0;
  char ans;
  while (v[v.size() - 1] - 'a' > i) {
    ans = 'a' + i;
    for (int j = 0; j < cnt[i]; ++j) cout << ans;
    ++i;
  }
  for (i = 0; i < v.size() - 1; ++i)
    if (v[i] == v[v.size() - 1]) break;
  while (i < v.size()) {
    cout << v[i];
    ++i;
  }
  printf("\n");
}
