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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
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 = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nextPrime(long long N) {
  if (N <= 1) return 2;
  long long prime = N;
  bool found = false;
  while (!found) {
    prime++;
    if (isPrime(prime)) found = true;
  }
  return prime;
}
long long fact(long long n) {
  if (n == 1) return 1;
  return n * fact(n - 1);
}
long long cl(long long n, long long d) { return (n + d - 1) / d; }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
void solve() {
  long long n, k, i, ptr = 0;
  cin >> n >> k;
  string a;
  cin >> a;
  set<char> s;
  for (i = 0; i < n; i++) {
    s.insert(a[i]);
  }
  if (s.size() == 1) {
    for (i = 0; i < n; i++) cout << a[i];
    for (i = 0; i < k - 1; i++) cout << *s.begin();
    return;
  }
  s.clear();
  string ss = a;
  for (i = 1; i < n; i++) {
    long long f = 1;
    for (long long k = i; k < n; k++) {
      if (a[k] != a[k - i]) {
        f = 0;
        break;
      }
    }
    if (f) {
      ss = "";
      for (long long k = n - i; k < n; k++) ss = ss + a[k];
      break;
    }
  }
  cout << a;
  if (ss.size() == 0)
    for (i = 1; i < k; i++) cout << a;
  else
    for (i = 1; i < k; i++) cout << ss;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}