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.
R=lambda:map(int,input().split())

n, L = R()

L2 = "{0:b}".format(L)

t = len(L2)

c = list(R())

b = [0]*max(n, t)
l = [0]*max(n, t)

b[0] = float('inf')
temp = 0
for i in range(n-1, -1, -1):
    if c[i] < b[0]:
        b[0] = c[i]
        temp = i
    l[0] = 2 ** temp
    
for i in range(1, n):
    if l[i - 1] >= 2 ** i:
        b[i] = b[i - 1]
        l[i] = l[i - 1]
    else:
        if min(2*b[i-1], min(c[i:])) == 2*b[i-1]:
            b[i] = 2 * b[i - 1]
            l[i] = 2 * l[i - 1]
        
        else:
            temp = 0
            b[i] = float('inf')
            for j in range(n - 1 , i - 1 , -1):
                if c[j] < b[i]:
                    b[i] = c[j]
                    temp = i
            l[i] = 2**temp

for i in range(n, t):
    b[i] = 2 * b[i - 1]
    l[i] = 2 * l[i - 1]


def solve(t, price):
    if t <= 0:
        return price
    t2 = "{0:b}".format(t)
    try :return min(solve( t - l[len(t2) - 1], price + b[len(t2) - 1]),
               solve( t - l[len(t2)], price + b[len(t2)]))
    except: return (solve( t - l[len(t2) - 1], price + b[len(t2) - 1]))

print(solve(L, 0))