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;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
template <typename T>
inline std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
  bool first = true;
  os << "[";
  for (auto i : v) {
    if (!first) os << ", ";
    os << i;
    first = false;
  }
  return os << "]";
}
const int N = 2 * 1e5 + 10;
int n, x, y, q, k, m;
int a[N];
string s, t;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  cin >> t;
  s = "";
  int ans = -1;
  for (int i = 1; i < n; ++i) {
    int f = 0;
    int x = i;
    for (int j = 0; j < n; ++j) {
      if (t[x++] != t[j]) {
        f = 1;
        break;
      }
      if (x == n) break;
    }
    if (f == 0) {
      ans = i;
      break;
    }
  }
  if (ans == -1) {
    for (int i = 0; i < k; ++i) {
      cout << t;
    }
    return 0;
  }
  string ls = t.substr(n - ans);
  cout << t;
  for (int i = 0; i < k - 1; ++i) {
    cout << ls;
  }
  return 0;
}