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;
const long long M = 1e9 + 7;
const double pi = acos(-1);
int longstring(string s) {
  int n = s.length();
  if (n < 2) {
    return 0;
  }
  int len = 0;
  int i = 1;
  while (i < n) {
    if (s[i] == s[len]) {
      ++len;
      ++i;
    } else {
      i = i - len + 1;
      len = 0;
    }
  }
  return len;
}
int main() {
  long long n, k;
  cin >> n >> k;
  string t;
  cin >> t;
  long long flag = 0;
  set<char> tt;
  for (int i = 0; i < n; i++) tt.insert(t[i]);
  if (tt.size() == 1) {
    string s = t;
    for (int i = 1; i < k; i++) s += t[0];
    cout << s << endl;
    return 0;
  }
  long long ans = longstring(t);
  string s1 = "";
  ans = n - ans;
  long long i = n - 1;
  while (ans > 0) {
    s1 += t[i];
    ans--;
    i--;
  }
  reverse(s1.begin(), s1.end());
  string s = t;
  for (int i = 1; i < k; i++) s += s1;
  cout << s << endl;
  return 0;
}