A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.

You want to buy at least L liters of lemonade. How many roubles do you have to spend?

Input

The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.

Output

Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

Examples

Input

4 12
20 30 70 90


Output

150


Input

4 3
10000 1000 100 10


Output

10


Input

4 3
10 100 1000 10000


Output

30


Input

5 787787787
123456789 234567890 345678901 456789012 987654321


Output

44981600785557577

Note

In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.

In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.

In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.
n, m = map(int, raw_input().split())
a = map(int, raw_input().split())
b = []
cost = {}
for i in xrange(n):
    cost[1 << i] = a[i]
for i in xrange(n, 40):
    cost[1 << i] = 10**19

for i in xrange(n):
    for j in xrange(i, 40):
        cost[1 << j] = min(cost[1 << j], a[i] * (1 << (j - i)))

cache = {}

def solve(pos, rem):
    if(rem <= 0):
        return 0
    if(pos == 0):
        return cost[1] * rem
    try:
        tmp = cache[(pos, rem)]
        return tmp
    except KeyError:
        pass
    res = solve(pos - 1, rem)
    if(1 << pos >= rem):
        res = min(res, cost[1 << pos])
    else:
        req = rem / (1 << pos)
        if(m % (1 << pos) != 0):
            res = min(res, (req + 1) * cost[1 << pos])
        res = min(res, (req * cost[1 << pos]) + solve(pos - 1, rem - (req * (1 << pos))))
    cache[(pos, rem)] = res
    return res

print solve(39, m)
exit(0)
i = 39
res = 1e19
while(i >= 0):
    if(1 << i >= m):
        res = min(res, cost[1 << i])
    else:
        req = m / (1 << i)
        if(m % (1 << i) != 0):
            req += 1
        res = min(res, req * cost[1 << i])
    i -= 1
print res

for i in xrange(n):
    b.append([cost[1 << i], 1 << i])
b.sort()

def check(ccost, v):
    i = len(b) - 1
    p = 0
    while(i >= 0):
        if(b[i][0] > ccost):
            i -= 1
        #elif(b[i][0] < ccost):
        #    i -= 1
        else:
            q = ccost / b[i][0]
            assert(q * b[i][0] <= ccost)
            ccost -= (q * b[i][0])
            p += q * b[i][1]
    return p >= v

s = 1
e = 10**19
iter = 100
mid = 0
while(iter):
    iter-=1
    mid = (s + e) / 2
    if(check(mid, m)):
        e = mid
    else:
        s = mid

s = mid - 5;
if(s <= 1):
     s = 1;
e = mid + 5;
if(e >= (10**19)):
     e = 10**19;


while(s <= e):
    if(check(s, m)):
        print min(res, s)
        exit(0)
    s += 1