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;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
void fast() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
string t;
bool solve(int m) { return t.substr(0, m) == t.substr(t.size() - m); }
int main() {
  fast();
  ;
  int n, k;
  cin >> n >> k;
  cin >> t;
  int idx = 1;
  int mx = 0;
  while (idx <= t.size() / 2) {
    if (solve(idx)) {
      mx = max(mx, idx);
    }
    idx++;
  }
  string s = t.substr(mx);
  string a = t;
  for (int i = 0; i < k - 1; i++) a += s;
  mx = 0;
  while (k) {
    for (idx = 0; k; idx++) {
      if (a.substr(idx, t.size()) == t) {
        k--;
        mx = idx + t.size();
        if (k == 0) break;
      }
    }
  }
  cout << a.substr(0, mx) << "\n";
}