One day Vasya got hold of information on the Martian dollar course in bourles for the next *n* days. The buying prices and the selling prices for one dollar on day *i* are the same and are equal to *a*<sub class="lower-index">*i*</sub>. Vasya has *b* bourles. He can buy a certain number of dollars and then sell it no more than once in *n* days. According to Martian laws, one can buy only an integer number of dollars. Which maximal sum of money in bourles can Vasya get by the end of day *n*?
The first line contains two integers *n* and *b* (1 ≤ *n*, *b* ≤ 2000) — the number of days and the initial number of money in bourles. The next line contains *n* integers *a*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 2000) — the prices of Martian dollars.
from collections import deque, Counter, OrderedDict
from heapq import nsmallest, nlargest

def binNumber(n,size):
    return bin(n)[2:].zfill(size)

def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a,a)

def iar():
    return list(map(int,input().split()))

def ini():
    return int(input())

def isp():
    return map(int,input().split())

def sti():
    return str(input())

def rln(a):
    return range(len(a))

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#      |       /____\      |    |  /  |
#      |      /      \     |    | /   |
#  ========= /        \  =====  |/====|  
#  code
if __name__ == "__main__":
    n,b = isp()
    a = iar()
    ma = b
    for i in range(n-1):
        for j in range(i+1,n):
            buy = b//a[i]
            rem = b%a[i]
            sel = buy*a[j] + rem
            ma = max(ma,sel)
    print(ma)
