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;
int INF = (int)1e9;
long long INFINF = (long long)1e18 + 10;
const long double PI = 3.14159265358979323846;
long long powermodm(long long x, long long n, long long mod) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return result;
}
long long GCD(long long, long long);
long long LCM(long long, long long);
long long power(int, int);
long long choose(long long, long long);
int ones(long long);
void extendedEuclid(long long, long long);
long long MMI(long long, long long);
void fastscan(int &);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int ans = 0;
  for (int i = 1; i < n; i++) {
    int c = 0;
    for (int j = 0; j <= i; j++) {
      if (s[j] == s[n - i + j]) {
        c++;
      }
    }
    if (c == i) {
      ans = i;
    }
  }
  if (k == 1) {
    cout << s << endl;
  } else {
    string st = "";
    for (long long i = ans; i < n; i++) {
      st = st + s[i];
    }
    cout << s;
    for (long long i = 0; i < k - 1; i++) {
      cout << st;
    }
    cout << endl;
  }
  return 0;
}
long long GCD(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
  return (max(a, b) / GCD(a, b)) * min(a, b);
}
long long power(int a, int n) {
  unsigned long long int result = 1, x = a;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
int ones(long long n) {
  int c = 0;
  while (n) {
    n = n & (n - 1);
    c++;
  }
  return c;
}
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long MMI(long long a, long long p) {
  extendedEuclid(a, p);
  if (d == 1 && p != 1)
    return ((x % p) + p) % p;
  else
    return -1;
}
void fastscan(int &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}