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;
void solve() {
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  map<char, long long> m;
  for (__typeof((n)) i = (0); i < (n); i++) m[s[i]]++;
  if ((int)m.size() == 1) {
    long long c = k - 1;
    while (c--) s += s[0];
    {
      cout << s << endl;
      return;
    };
  }
  string t, g;
  long long x = -1;
  for (__typeof((n - 1)) i = (0); i < (n - 1); i++) {
    if (s.substr(0, i + 1) == s.substr(n - i - 1, i + 1)) x = i;
  }
  if (x == -1) {
    while (k--) {
      t += s;
    }
    {
      cout << t << endl;
      return;
    };
  }
  string ans;
  t = s.substr(0, n - x - 1);
  g = s.substr(n - 1 - x, x + 1);
  while (k--) ans += t;
  ans = ans + g;
  cout << ans << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}