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
n, l = list(map(int, input().split()))
arr = list(map(int, input().split()))
dp = [0 for i in range(32)]

for i in range(32):
    dp[i] = (2 ** i) * arr[0]

for i in range(1, n):
    for j in range(i, 32):
        f = 2 ** j
        cur = 2 ** i
        fac = f // cur
        dp[j] = min(dp[j], fac * arr[i])

temp = []
for i in range(n):
    temp.append([dp[i] / (2 ** i), 2 ** i, dp[i]])
temp.sort()


def checker(cost):
    cur_vol = 0
    while True:
        f = 0
        for i in range(n):
            if cost >= temp[i][2]:
                fac = cost // temp[i][2]
                num = fac*temp[i][1]
                cur_vol += num
                f = 1
                cost = cost - (fac * temp[i][2])
                break
        if f == 0:
            break
    if cur_vol >= l:
        return 1
    else:
        return - 1


low = 0
high = 10 ** 19
while low < high:
    mid = (low + high) // 2
    f = checker(mid)
    if f == 1:
        high = mid
    else:
        low = mid
    if high - low == 1:
        break
if checker(low) == 1:
    print(low)
else:
    print(high)
