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 N = 1e6 + 10;
int n, k, r, q;
string x, y, z;
int main() {
  cin >> n >> k >> x;
  for (int i = 0; i < n / 2; i++) {
    if (x[r] == x[(n + 1) / 2 + i])
      r++;
    else {
      i -= r;
      r = 0;
    }
    if (i == (n / 2) - 1 && x[r - 1] != x[(n + 1) / 2 + i]) r = 0;
  }
  for (int i = 0; i < n / 2; i++) {
    if (x[i] == x[n - 1 - i] && x[i] == x[0])
      q++;
    else
      break;
  }
  if (r == 0) r = q;
  y = x;
  if (r != 0) {
    for (int i = r; i < n; i++) {
      y += x[i];
      z += x[i];
      if (y.find(x, y.find(x, 0) + 1) != -1) break;
    }
    cout << x;
    for (int i = 0; i < k - 1; i++) cout << z;
  } else {
    for (int i = 0; i < k; i++) cout << x;
  }
}