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 int maxn = 1e4 + 5;
int nex[maxn];
void getnex(string s) {
  int n = s.size(), i = 0, j = -1;
  nex[0] = -1;
  while (i < n) {
    if (s[i] == s[j] || j == -1)
      nex[++i] = ++j;
    else
      j = nex[j];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  getnex(s);
  int pos = nex[n];
  string t = s;
  k--;
  while (k--) {
    for (int i = pos; i < n; i++) t += s[i];
  }
  cout << t << '\n';
  return 0;
}