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.
n,p=map(int,input().split())
l=list(map(int,input().split()))
start=None
end=None
new=[]
for i in range(n-1):
    if l[i+1]>l[i] and start==None:
        start=i
    elif l[i+1]<l[i] and start!=None:
        end = i
        new.append(l[start:end+1])
        start=None
        end=None

if start!=None:
    new.append(l[start:])
if len(new)==0:
    print(p)
    exit()
ans=0
dollars=0
maxex=[]
mins=[]
for i in new: 
    if i[0]<p:
        mins.append(i[0])
        maxex.append(i[-1])
        '''dollars+=p//i[0]
        print("dollars and i",dollars,i)
        p-=i[0]*dollars
        print("old p",p)
        p+=i[-1]*dollars
        print("new p",p)
        dollars=0
        print("here",p,i[0],i[-1])'''
oldp=p
record=[]
for i in range(len(mins)):
    p=oldp
    dollars=p//mins[i]
    p-=dollars*mins[i]
    p+=dollars*(max(maxex[i:]))
    record.append(p)


print(max(record))
#print(p)