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 int N = 1e4 + 110;
const int MAXN = 1100;
const int inf = 0x7fffffff;
const double eps = 1e-8;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, k;
string s;
int len1, len2, next1[N];
char S[N], T[N];
inline void get_next() {
  int t1 = 0, t2;
  next1[0] = t2 = -1;
  while (t1 < len2) {
    if (t2 == -1 || T[t1] == T[t2])
      next1[++t1] = ++t2;
    else
      t2 = next1[t2];
  }
}
inline void KMP() {
  int t1 = 0, t2 = 0;
  while (t1 < len1) {
    if (t2 == -1 || S[t1] == T[t2])
      t1++, t2++;
    else
      t2 = next1[t2];
    if (t2 == len2) {
      printf("%d\n", t1 - len2 + 1);
      t2 = next1[t2];
    }
  }
}
int main() {
  n = read(), k = read();
  cin >> s;
  int len = s.size();
  n = len;
  int pos = 0;
  for (register int i = 1; i < len; i++)
    if (s.substr(0, i) == s.substr(n - i, i)) pos = i;
  for (register int j = 1; j < k; j++)
    for (register int i = pos; i < len; i++) s = s + s[i];
  cout << s << endl;
  return 0;
}