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 long long maxx = 1e9 + 7;
bool compare(const pair<long long, long long> &a,
             const pair<long long, long long> &b) {
  return a.first > b.first;
}
long long inline power(long long a, long long b, long long p) {
  a %= p;
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
long long inv(long long n, long long p) { return power(n, p - 2, p); }
bool inline isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
template <class T>
void printv(T &v, long long n) {
  if (n == -1) n = v.size();
  for (long long i = 0; i < n; i++) cout << v[i] << " ";
  cout << "\n";
}
void solve() {
  long long n, k, idx;
  cin >> n >> k;
  idx = n;
  string t;
  cin >> t;
  string ans = "";
  for (long long i = 1; i <= n - 1; i++) {
    if (t.substr(i) == t.substr(0, n - i)) {
      idx = i;
      break;
    }
  }
  for (long long i = 0; i < k; i++) {
    ans += t.substr(0, idx);
  }
  if (idx < n) ans += t.substr(idx);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 1;
  while (test--) {
    solve();
  }
}