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 double pi = 3.1415926535897932384626433832795;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int fast_pow(long long int a, long long int b,
                              long long int mod) {
  long long int res = 1;
  a %= mod;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
inline bool isPrime(long long int a) {
  for (long long int i = 2; i * i <= a; i++)
    if (a % i == 0) return 0;
  return 1;
}
string t, s, temp = "";
int k, n, pos, flag;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> k >> t;
  for (int i = 1; i < n; i++) {
    flag = true;
    for (int j = 0, l = i; l < n; j++, l++) {
      if (t[j] != t[l]) {
        flag = false;
        break;
      }
    }
    if (flag) {
      pos = n - i;
      break;
    }
  }
  s = t;
  if (pos) {
    for (int i = pos; i < n; i++) temp += t[i];
    for (int i = 1; i < k; i++) s += temp;
  } else {
    for (int i = 1; i < k; i++) s += t;
  }
  cout << s << '\n';
  return 0;
}