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};
int dy[] = {1, 0, -1, 0};
int ddx[] = {1, 0};
int ddy[] = {1, 1};
int n, k;
string s;
string ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  cin >> s;
  if (n == 1) {
    for (int i = 0; i < k; i++) ans += s;
    cout << ans << "\n";
    return 0;
  } else if (k == 1) {
    cout << s << "\n";
    return 0;
  }
  int tam = 0;
  ans = s;
  string aux = "";
  for (int i = 1; i < int(ans.size()); i++) {
    bool at = false;
    int aa = 0;
    for (int j = 0; j < int(s.size()); j++) {
      if (i + j >= int(ans.size())) break;
      if (ans[i + j] != s[j]) {
        at = true;
        break;
      } else
        aa++;
    }
    if (!at) tam = max(aa, tam);
  }
  aux = s.substr(tam, int(s.size()));
  for (int i = 1; i < k; i++) ans += aux;
  cout << ans << "\n";
  return 0;
}