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 INF = 1000000000 + 7;
const double esp = 1e-13;
const long double PI = acos(-1.0);
long long nhan(long long x, long long y, long long m) {
  long long ans = 0;
  while (y) {
    if (y % 2) ans = (ans + x) % m;
    x = x * 2 % m;
    y /= 2;
  }
  return ans;
}
long long power(long long a, long long n, long long m) {
  unsigned long long ans = 1;
  while (n) {
    if (n % 2) ans = ans * a % m;
    a = a * a % m;
    n /= 2;
  }
  return ans;
}
bool check(long long n) {
  if (n == 2) return true;
  if (n < 2 || n % 2 == 0) return false;
  long long p[3] = {3, 5, 7};
  long long a, d;
  int r, s;
  d = n - 1;
  s = 0;
  while (d % 2 == 0) {
    d /= 2;
    s++;
  }
  for (long long i = 0; i <= 2; i++) {
    if (n == p[i]) return true;
    if (n % p[i] == 0) return false;
    a = power(p[i], d, n);
    if (a != 1) {
      for (r = 0; r < s && a != n - 1; r++) a = a * a % n;
      if (r == s) return false;
    }
  }
  return true;
}
long long phi(long long n) {
  long long ans = n;
  long long n2 = (long long)sqrt(n);
  for (long long i = 2; i <= n2; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      ans -= ans / i;
    }
  if (n > 1) ans -= ans / n;
  return ans;
}
string s;
int n, k, cnt;
char s2[3000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n >> k;
  cin >> s;
  for (long long i = 0; i < n * k; i++) s2[i] = ' ';
  cnt = 0;
  for (long long i = 0; i < n * k; i++) {
    int x = 0;
    int j = i;
    while (j < n * k && x < s.length()) {
      if (s2[j] == s[x]) {
        j++;
        x++;
      } else
        break;
    }
    if (s2[j] == ' ') {
      cnt++;
      while (x < s.length()) {
        s2[j] = s[x];
        x++;
        j++;
      }
      if (cnt == k) break;
    }
  }
  for (long long i = 0; i < n * k; i++)
    if (s2[i] != ' ') cout << s2[i];
}