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 LLMAX = 2e18;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
template <class T>
inline void read(T &DataIn) {
  DataIn = 0;
  T Flag = 0;
  char c = getchar();
  while (!isdigit(c)) {
    Flag |= c == '-';
    c = getchar();
  }
  while (isdigit(c)) {
    DataIn = DataIn * 10 + c - '0';
    c = getchar();
  }
  DataIn = Flag ? -DataIn : DataIn;
}
template <class T>
inline void write(T DataOut, char EndChar = '\n') {
  T lenth = 0, number[30];
  if (DataOut == 0) {
    putchar(48);
    return;
  }
  while (DataOut > 0) {
    number[++lenth] = DataOut % 10;
    DataOut /= 10;
  }
  for (int i = lenth; i >= 1; i--) putchar(number[i] + 48);
  putchar(EndChar);
}
priority_queue<int, vector<int>, less<int> > qd;
priority_queue<int, vector<int>, greater<int> > qu;
int main(void) {
  FILE *fin = NULL, *fout = NULL;
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s, cpy;
  int n, k, len = 0;
  cin >> n >> k >> s;
  for (int j = 1; j < s.size(); j++) {
    int flag = 1;
    for (int x = 0, y = j; y < s.size(); x++, y++)
      if (s[x] != s[y]) flag = 0;
    if (flag == 1) {
      for (int i = s.size() - j; i < s.size(); i++) cpy += s[i];
      cout << s;
      for (int i = 1; i < k; i++) cout << cpy;
      return 0;
    }
  }
  for (int i = 1; i <= k; i++) cout << s;
  return 0;
}