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.
import math

def remove_items(a, removed):
    return [a[i] for i in range(len(a)) if i not in removed]

def refine1(c, v):
    removed = set()
    for i in range(len(c) - 1):
        if i in removed:
            continue
        for j in range(i + 1, len(c)):
            if j in removed:
                continue
            if c[i] / v[i] < c[j] / v[j]:
                removed.add(j)
    return remove_items(c, removed), remove_items(v, removed)

def refine2(c, v):
    removed = set()
    for i in range(len(c)):
        for j in range(i + 1, len(c)):
            if c[i] >= c[j]:
                removed.add(i)

    return  remove_items(c, removed), remove_items(v, removed)

def result(L, c, v, cutoff):
    ans = 0
    while L > 0:
        if cutoff == 0:
              num = math.ceil(L / v[-1])
              ans += num * c[-1]
              return ans
        num = L // v[-1]
        L -= num * v[-1]
        ans += num * c[-1]
        cutoff -= 1
        v.pop()
        c.pop()
    return ans

n, L = map(int, input().split())
c = list(map(int, input().split()))
v = [2 ** i for i in range(n)]

c, v = refine1(c, v)
c, v = refine2(c, v)
# print(c)
# print(v)

ans = float('inf')
for cutoff in range(len(c)):
    ans = min(ans, result(L, c[:], v[:], cutoff))

print(ans)    
