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:
def counter(s, t): ## O(n**2) potentially oops

    how_many = 0

    for i in range(0, len(s)-len(t)+1):

        ## print(i, i+len(t))

        if s[i:i+len(t)] == t:
            how_many += 1

    return how_many

while True:
    
    inp = input().split()

    n = int(inp[0])
    k = int(inp[1])

    s = input()

    returning = s

    for i in range(1, n+1):
        if counter(returning+s[-i:], s) > counter(returning, s): ## (returning + s[-i:]).count(s) > returning.count(s):
            addition = s[-i:]
            break

    returning = s + (k-1)*addition

##    while counter(returning, s) < k: ## returning.count(s) < n:
##
##        for i in range(1, n+1):
##
##            if counter(returning+s[-i:], s) > counter(returning, s): ## (returning + s[-i:]).count(s) > returning.count(s):
##
##                returning += s[-i:]
##                break

    print(returning)

    break