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, l = map(int, input().split())
c = list(map(int, input().split()))

v = []
for i in range(n):
    v.append(2**i)
from math import ceil
def sl(n,l,c, bl):
    #print(n, l, c, bl)
    s = []
    for i in range(n):
        if i not in bl:
            s.append((v[i], v[i]/c[i], c[i], i))
    sm = min(s, key=lambda x: -x[1])
    ct = ceil(l / sm[0])
    if sm[0] == 1:
        return ct*sm[2]
    ans = (ct-1)*sm[2]
    ans1 = min(sm[2], sl(n, l-(ct-1)*sm[0], c, bl + [sm[3]]))
    return ans + ans1
print(sl(n, l, c, []))