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.
# Problemset 41 B

#we want to buy on days where dollars cost min
#sell when dollars cost max
# ==> we have the maximum number of bourles

# if you buy today you can't sell yesterday


#p tuple: we want to know what is the maximum diff between the price of dollars.
def max_diff(tpl):
    md_f = 0, 1

    for i in range(1, len(tpl)):
        for j in range(i+1, len(tpl)):
            if tpl[j] - tpl[i] > tpl[md_f[1]] - tpl[md_f[0]]:
                md_f = i, j

    return md_f if tpl[md_f[1]] - tpl[md_f[0]] > 0 else -1


n, b = map(int, input().split())  #n = days // b = number of bourles
p = tuple(map(int, input().split()))  #prices per day for n days

md = max_diff(p)

if md == -1:
    print(b)
else:
    print(divmod(b, p[md[0]])[1] + divmod(b, p[md[0]])[0]*p[md[1]])

