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;
vector<int> getPrefixFunc(string str) {
  vector<int> prefix;
  prefix.push_back(0);
  int j;
  int flag;
  for (int i = 1; i < str.length(); i++) {
    flag = 0;
    j = prefix[i - 1];
    while (j > 0) {
      if (str[j] == str[i]) {
        prefix.push_back(j + 1);
        flag = 1;
        break;
      }
      j = prefix[j - 1];
    }
    if (flag == 0) {
      if (str[0] == str[i]) {
        prefix.push_back(1);
      } else {
        prefix.push_back(0);
      }
    }
  }
  return prefix;
}
int main() {
  int n, k;
  string pattern;
  cin >> n;
  cin >> k;
  cin >> pattern;
  vector<int> prefix;
  prefix = getPrefixFunc(pattern);
  string ans = pattern;
  int leftCharacter = pattern.length() - prefix[pattern.length() - 1];
  for (int add = 0; add < k - 1; add++) {
    for (int i = prefix[pattern.length() - 1]; i < pattern.length(); i++) {
      ans += pattern[i];
    }
  }
  cout << ans << endl;
  return 0;
}