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:
import math
n, k = map(int, input().split())
s = input()
j = -1
cnt = 0
for i in range(1, n + 1):
    if i > (n - i):
        break
    if s[:i] == s[n - i:]:
        cnt = i
otv = ''
for i in range(k):
    if i % 2 == 0:
        otv += s
    else:
        otv += s[cnt:n - cnt]
        
if s.count(s[0]) == n:
    print(s[0] * k + s[0] * (n - 1))
    exit()
    
if k % 2 == 0:
    otv += s[:cnt]
    cnt = 0
    for i in range(len(otv)):
        if otv[i: i + n] == s:
            cnt += 1
        if cnt == k:
            print(otv[:i + n])
            break
else:
    cnt = 0
    for i in range(len(otv)):
        if otv[i: i + n] == s:
            cnt += 1
        if cnt == k:
            print(otv[:i + n])
            break