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.
from math import ceil, floor

n,L = map(int, input().split(" "))
cost = list(map(int, input().split(" ")))
cost.extend([-1 for _ in range(30-len(cost))])

for i in range(1,30):
    if 2*cost[i-1] < cost[i] or cost[i] == -1:
        cost[i] = 2 * cost[i-1]

for i in range(29,1,-1):
    if cost[i-1] > cost[i]:
        cost[i-1] = cost[i]

totalCost = 0
altTotalCost = 100000000000000000000000
while L > 0:

    index1 = L.bit_length()-1
    index2 = L.bit_length()

    if index2 >= 30:
        #fill up
        num = floor(L / 2**index1)
        totalCost += cost[index1] * num
        L -= num * 2**index1
        continue

    L -= 2**index1
    totalCost += cost[index1]
    altTotalCost = min(altTotalCost, totalCost - cost[index1] + cost[index2])
    
print(min(totalCost, altTotalCost))
