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.
n, L = map(int, input().split()); P = L
def log(n):
    a = 1
    while a * 2 <= n:
        a *= 2
    return a
c = list(map(int, input().split()))
for i in range(1, n):
    c[i] = min(c[i - 1] * 2, c[i])
for i in range(n - 2, -1, -1):
    c[i] = min(c[i], c[i + 1])
Mark = list(map(int, ("0 " * (n)).split()))
d = len(c) - 2
a = 2**d
ans = c[-1] * (L // 2**(n - 1))
L %= 2**(n - 1)
while L != 0:
    while L < a:
        a //= 2
        d -= 1
    Mark[d] = 1
    L -= a
    ans += c[d]
P = 0
for i in range(len(c)):
    if Mark[i] == 1:
        P += c[i]
        continue
    ans -= max(0, P - c[i])
    if P - c[i] > 0:
        P = c[i]
print(ans)

