Problem: You are given a string t consisting of n lowercase Latin letters and an integer number k.

Let's define a substring of some string s with indices from l to r as s[l ... r].

Your task is to construct such string s of minimum possible length that there are exactly k positions i such that s[i ... i + n - 1] = t. In other words, your task is to construct such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 50) — the length of the string t and the number of substrings.

The second line of the input contains the string t consisting of exactly n lowercase Latin letters.

Output

Print such string s of minimum possible length that there are exactly k substrings of s equal to t.

It is guaranteed that the answer is always unique.

Examples

Input

3 4
aba


Output

ababababa


Input

3 2
cat


Output

catcat

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void pisz(int n) { printf("%d\n", n); }
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
int main() {
  long long n, k;
  cin >> n >> k;
  std::vector<char> t(n, 0);
  for (size_t i = 0; i < n; i++) {
    cin >> t[i];
  }
  long long index = 0;
  for (size_t i = 1; i < n; i++) {
    if (t[0] == t[i]) {
      long long k = 1;
      int flag = 1;
      for (size_t j = i + 1; j < n; j++) {
        if (t[j] != t[k]) {
          flag = 0;
          break;
        } else {
          k++;
        }
      }
      if (flag == 1) {
        index = i;
        break;
      }
    }
  }
  if (index == 0) {
    for (size_t i = 0; i < k; i++) {
      for (size_t j = 0; j < n; j++) {
        cout << t[j];
      }
    }
    return 0;
  }
  index = n - 1 - index;
  for (size_t i = 0; i < n; i++) {
    cout << t[i];
  }
  for (size_t i = 0; i < k - 1; i++) {
    for (size_t j = index + 1; j < n; j++) {
      cout << t[j];
    }
  }
  return 0;
}