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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
void solve(istream& in, ostream& out) {
  int n, k;
  string t;
  in >> n >> k >> t;
  string l = "", r = "";
  vector<int> ind;
  for (int i = (0); i < (n); i++) {
    l += t[i];
    r = t[n - i - 1] + r;
    if (l == r) {
      ind.push_back(i);
    }
  }
  ind.pop_back();
  if (ind.size() == 0) {
    for (int i = (0); i < (k); i++) out << t;
    return;
  }
  int idx = *ind.rbegin();
  string tt = "";
  for (int i = (idx + 1); i < (n); i++) {
    tt += t[i];
  }
  out << t;
  for (int i = (0); i < (k - 1); i++) out << tt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve(cin, cout);
  return 0;
}