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;
char s[1000005];
int vis[1000005], dp[1000005][30];
int tree[1000005] = {0}, n, INF = 0x3f3f3f3f;
void add(int x) {
  while (x <= n) {
    tree[x]++;
    x += x & (-x);
  }
}
int sum(int x) {
  int ans = 0;
  while (x) {
    ans += tree[x];
    x -= x & (-x);
  }
  return ans;
}
void makeRmqIndex(int n, char b[]) {
  int i, j;
  for (i = 0; i < n; i++) dp[i][0] = i;
  for (j = 1; (1 << j) <= n; j++)
    for (i = 0; i + (1 << j) - 1 < n; i++) {
      if (b[dp[i][j - 1]] < b[dp[i + (1 << (j - 1))][j - 1]])
        dp[i][j] = dp[i][j - 1];
      else if (b[dp[i][j - 1]] == b[dp[i + (1 << (j - 1))][j - 1]]) {
        dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
      } else
        dp[i][j] = dp[i + (1 << (j - 1))][j - 1];
    }
}
int rmqIndex(int s, int v, char b[]) {
  int k = (int)(log((v - s + 1) * 1.0) / log(2.0));
  if (b[dp[s][k]] < b[dp[v - (1 << k) + 1][k]]) {
    return dp[s][k];
  } else if (b[dp[s][k]] == b[dp[v - (1 << k) + 1][k]]) {
    return max(dp[s][k], dp[v - (1 << k) + 1][k]);
  } else
    return dp[v - (1 << k) + 1][k];
}
int main() {
  int m;
  scanf("%d%s", &m, s);
  memset(vis, 0, sizeof(vis));
  n = strlen(s);
  makeRmqIndex(n, s);
  char mn = 'z';
  int zz['a' + 30] = {0};
  for (int i = 0; i < n; i++) {
    if (s[i] < mn) mn = s[i];
    zz[s[i]]++;
  }
  char xx = mn;
  int num = 0;
  for (int i = 0; i < n - m + 1; i++) {
    if (i > 0 && sum(i + m - 1 + 1) - sum(i - 1 + 1) > 0) {
      ;
    } else {
      int x = rmqIndex(0 + i, i + m - 1, s);
      if (s[x] > xx)
        xx = s[x], num = 1;
      else if (s[x] == xx)
        num++;
      if (vis[x] == 0) {
        vis[x] = 1;
        add(x + 1);
      }
    }
  }
  for (char i = 'a'; i <= 'z' && i < xx; i++) {
    for (int j = 0; j < zz[i]; j++) putchar(i);
  }
  for (int i = 0; i < num; i++) {
    putchar(xx);
  }
  puts("");
}
