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;
bool 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 += 6) {
    if ((n % i == 0) || (n % (i + 2) == 0)) return false;
  }
  return true;
}
int32_t main() {
  long long n, k;
  cin >> n >> k;
  string s, raj = "";
  cin >> s;
  raj = s;
  long long hash[26] = {0}, mx = 0;
  for (long long i = 0; i < n; i++) {
    hash[s[i] - 'a']++;
    mx = max(mx, hash[s[i] - 'a']);
  }
  if (mx == n) {
    cout << s;
    for (long long i = 0; i < (k - 1); i++) {
      cout << s[0];
    }
    return 0;
  }
  long long c = 0, d = 0;
  for (long long i = 1; i < n; i++) {
    c = 0;
    if (raj[0] == s[i]) {
      for (long long j = i; j < n; j++) {
        if (raj[j - i] == s[j]) {
          c++;
        } else {
          c = 0;
          break;
        }
      }
    }
    d = max(d, c);
  }
  string z = "";
  for (long long i = d; i < n; i++) z += s[i];
  cout << s;
  for (long long i = 0; i < (k - 1); i++) cout << z;
}