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;
vector<int> vec[200];
int segID = 0;
int segX[100010];
int segY[100010];
int used[200];
vector<int> usedLetter;
int taken[100010];
int freq[200];
bool visited[100010];
int main() {
  memset(taken, 0, sizeof(taken));
  memset(freq, 0, sizeof(freq));
  memset(used, 0, sizeof(used));
  memset(visited, false, sizeof(visited));
  int m;
  cin >> m;
  string s;
  cin >> s;
  int len = s.size();
  set<int> st;
  for (int i = 0; i < len; i++) {
    int x = s[i];
    freq[x]++;
    vec[x].push_back(i);
    if (used[x] == 0) usedLetter.push_back(x), used[x] = 1;
    if (i + m - 1 < len) {
      st.insert(i);
      segX[segID] = i;
      segY[segID] = i + m - 1;
      segID++;
    }
  }
  sort(usedLetter.begin(), usedLetter.end());
  set<int>::iterator it;
  int mxI = 0;
  for (int k = 0; k < usedLetter.size() && !st.empty(); k++) {
    if (vec[usedLetter[k]].size() > 0) {
      bool found = false;
      int i = usedLetter[k];
      int savePrvFoundID = -1;
      for (int j = 0; j < segID && !st.empty(); j++) {
        int x = segX[j];
        int y = segY[j];
        if (savePrvFoundID != -1 && savePrvFoundID >= x &&
            savePrvFoundID <= y) {
          if (!visited[x]) {
            visited[x] = true;
            mxI = max(mxI, i);
          }
        } else if (!visited[x]) {
          int idY =
              lower_bound(vec[i].begin(), vec[i].end(), y) - vec[i].begin();
          if (vec[i][idY] != y) idY--;
          if (vec[i][idY] >= x && vec[i][idY] <= y && idY >= 0) {
            visited[x] = true;
            taken[vec[i][idY]] = 1;
            mxI = max(mxI, i);
            savePrvFoundID = vec[i][idY];
          }
        }
      }
    }
  }
  for (int i = 97; i < mxI; i++) {
    for (int j = 0; j < freq[i]; j++) {
      char c = i;
      cout << c;
    }
  }
  for (int i = 0; i < len; i++) {
    int x = s[i];
    if (x == mxI && taken[i] == 1) cout << s[i];
  }
  cout << "\n";
  return 0;
}
