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 MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
template <class T>
inline T gcd(T a, T b) {
  T r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
inline T fpow(T n, T k, int p = MOD) {
  T r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
const int MAXN = 100000 + 5;
int sum[MAXN], sum2[MAXN];
int cnt[30];
void input() {}
void solve() {
  int m;
  string str;
  cin >> m;
  cin >> str;
  for (int i = 0; i < 26; i++) {
    sum[0] = 0;
    for (int j = 1; j <= str.length(); j++) {
      sum[j] = sum[j - 1] + (str[j - 1] <= 'a' + i ? 1 : 0);
    }
    bool ok = true;
    for (int j = 1; j <= str.length() - m + 1; j++) {
      if (sum[j + m - 1] - sum[j - 1] == 0) {
        ok = false;
        break;
      }
    }
    if (ok) {
      memset(cnt, 0, sizeof(cnt));
      for (int j = 0; j < str.length(); j++)
        if (str[j] < 'a' + i) {
          cnt[str[j] - 'a']++;
        }
      sum[0] = sum2[0] = 0;
      for (int j = 1; j <= str.length(); j++) {
        sum[j] = sum[j - 1] + (str[j - 1] < 'a' + i ? 1 : 0);
        sum2[j] = sum2[j - 1] + (str[j - 1] == 'a' + i ? 1 : 0);
      }
      int choose = 0;
      for (int j = 1; j <= str.length(); j++) {
        if (choose > 0) {
          if (str[j - 1] - 'a' == i) {
            choose--;
            if (choose == 0) {
              cnt[i]++;
            }
          }
        } else if (j <= str.length() - m + 1 &&
                   sum[j + m - 1] - sum[j - 1] == 0) {
          choose = sum2[j + m - 1] - sum2[j - 1];
          if (str[j - 1] - 'a' == i) {
            choose--;
            if (choose == 0) {
              cnt[i]++;
            }
          }
        }
      }
      break;
    }
  }
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < cnt[i]; j++) {
      printf("%c", 'a' + i);
    }
  }
  printf("\n");
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  input();
  solve();
  return 0;
}
