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 itertools import permutations

MOD = 10**9+7

def find_best(c):
    n = 2 ** len(c)
    best_c = c[0] * n
    best_i = 0
    for i, ci in enumerate(c):
        if n // (2**i) * c[i] < best_c:
            best_c = n // (2**i) * c[i]
            best_i = i
    return best_i

def main():
    n, l = [int(c) for c in input().split(' ')]
    c = [int(c) for c in input().split(' ')]
    ans = []
    cur_ans = 0
    while l:
        ind = find_best(c[:n])
        vol = 2**ind
        cnt = l // vol
        l %= vol
        cur_ans += cnt * c[ind]
        ans.append(cur_ans + (l % vol and c[ind] or 0))
        n = ind
    print(min(ans))

while 1:
    main()
    break
# input()