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.
def one_case():
    n, ll = map(int, input().split())
    a = [int(x) for x in input().split()]
    if n == 1:
        print(a[0] * ll)
        return
    d = [2 ** i for i in range(n)]

    for i in range(1, n):
        if a[i] > a[i - 1] * 2:
            a[i] = a[i - 1] * 2
    for i in range(n - 2, -1, -1):
        if a[i] > a[i + 1]:
            a[i] = a[i + 1]
    # print(a)

    high = 0
    if ll > d[n - 1]:
        left = ll % d[n - 1]
        unit = (ll - left) // d[n - 1]
        high = unit * a[n - 1]
        ll = left

    def calc_cost(ll, i):
        # print(ll, i)
        if i == 0:
            return a[0] * ll
        base_cost = a[i + 1]
        try_cost = calc_cost(ll % d[i], i - 1) + a[i] * (ll // d[i])
        # print(ll, i, base_cost, try_cost)
        return min(base_cost, try_cost)

    print(calc_cost(ll, n - 2) + high)


# t = int(input())
t = 1
for i in range(t):
    one_case()
