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 PYTHON3):

Solution:
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 18/07/2020

from sys import stdin,stdout
from math import gcd, ceil, sqrt
from collections import Counter
from bisect import bisect_left, bisect_right
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007

def lps(s):
    arr = [0] * (len(s))
    i, j = 1, 0
    while i < len(s):
        if s[i] == s[j]:
            arr[i] = j + 1
            j += 1
        else:
            while j > 0 and s[i] != s[j]:
                j = arr[j - 1]
            if s[i] == s[j]:
                arr[i] = j + 1 
                j += 1
        i += 1
    return arr

n, k = iia()
t = is1()
common = lps(t)[-1]
res = t
for i in range(k - 1):
    res += t[common:]
print(res)