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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
const long long inf = 1e18;
const long long minf = -(1e18);
const long long mod = 1e9 + 7;
const long long med = 1e9 + 9;
inline long long add(long long x, long long y) {
  x %= mod;
  y %= mod;
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long sub(long long x, long long y) {
  x %= mod;
  y %= mod;
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long mul(long long x, long long y) {
  x %= mod;
  y %= mod;
  return (x * 1ll * y) % mod;
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long ans = power(x, y / 2) % mod;
  ans *= ans;
  ans %= mod;
  if (y % 2) ans *= x;
  return ans % mod;
}
long long inv(long long x, long long y) {
  x += mod;
  y += mod;
  x %= mod;
  y %= mod;
  long long ret = (power(y, mod - 2) % mod) * x;
  return (ret % mod);
}
bool check(string a, string b) {}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  string t;
  cin >> t;
  int l = t.length();
  int x = t[0];
  if (l == 1) {
    for (int i = 1; i <= k; i++) cout << t[0];
    return 0;
  }
  for (int i = 1; i < t.length(); i++) {
    if (t[i] != x)
      break;
    else if (i == t.length() - 1) {
      cout << t;
      for (int i = 1; i <= k - 1; i++) cout << t[0];
      return 0;
    }
  }
  int counter = 0;
  string temp;
  for (int i = l - 2; i >= 0; i--) {
    if (t.substr(l - 1 - i, i + 1) != t.substr(0, i + 1)) {
      temp = t.substr(i, l - i);
    } else {
      break;
    }
  }
  cout << t;
  for (int j = 1; j <= k - 1; j++) {
    cout << temp;
  }
  return 0;
  return 0;
}