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;
template <class A>
void pr(A a) {
  cout << a;
  cout << '\n';
}
template <class A, class B>
void pr(A a, B b) {
  cout << a << ' ';
  pr(b);
}
template <class A, class B, class C>
void pr(A a, B b, C c) {
  cout << a << ' ';
  pr(b, c);
}
template <class A, class B, class C, class D>
void pr(A a, B b, C c, D d) {
  cout << a << ' ';
  pr(b, c, d);
}
template <class A>
void PR(A a, long long n) {
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (i) cout << ' ';
    cout << a[i];
  }
  cout << '\n';
}
long long check(long long n, long long m, long long x, long long y) {
  return x >= 0 && x < n && y >= 0 && y < m;
}
const long long MAX = 1e9 + 7, MAXL = 1LL << 61, dx[4] = {-1, 0, 1, 0},
                dy[4] = {0, 1, 0, -1};
void Main() {
  long long n, m;
  string s;
  cin >> n >> m >> s;
  string t = "";
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    if (s.substr(0, i) == s.substr(n - i)) t = s.substr(i);
  }
  for (long long i = (long long)(0); i < (long long)(m - 1); i++) s += t;
  pr(s);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Main();
  return 0;
}