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;
inline int Int() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long Long() {
  long x;
  scanf("%ld", &x);
  return x;
}
inline long long LLong() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline float Float() {
  float x;
  scanf("%f", &x);
  return x;
}
inline double Double() {
  double x;
  scanf("%lf", &x);
  return x;
}
inline pair<int, int> Pair() {
  pair<int, int> p;
  scanf("%d %d", &p.first, &p.second);
  return p;
}
void Char(char *x) { scanf("%s", x); }
void pI(int x) { printf("%d", x); }
void pL(long x) { printf("%ld", x); }
void pLL(long long x) { printf("%lld", x); }
void pUL(unsigned long long x) { printf("%llu", x); }
void pF(float x) { printf("%f", x); }
void pD(double x) { printf("%f", x); }
void pS(char x[]) { printf("%s", x); }
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, 1, -1, 0, 0, -1, 1};
const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9 + 7;
const int maxn = (int)2e5 + 5;
const int LOGN = 20;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  string t;
  int n, k;
  cin >> n >> k;
  cin >> t;
  string s = t;
  for (int i = 1; i <= (k - 1); i++) {
    bool d = 0;
    for (int j = s.length() - n + 1; j < s.length(); j++) {
      bool ok = 1;
      int p = 0;
      for (int c = j; c < s.length(); c++) {
        if (s[c] != t[p]) {
          ok = 0;
          break;
        }
        ++p;
      }
      if (ok) {
        while (p < n) {
          s += t[p];
          ++p;
        }
        d = 1;
        break;
      }
    }
    if (d == 0) {
      int p = 0;
      while (p < n) {
        s += t[p];
        ++p;
      }
    }
  }
  cout << s;
  return 0;
}