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.
# tutorial codeforces :(
# مخم گوزید ناموسا
n, l = map(int, input().split())
c = list(map(int, input().split()))
res = 0
for i in range(1, n):
    c[i] = min(c[i - 1]*2, c[i])
p = 1 << (n - 1) # 2 ** (n - 1)
y = l // p
l-= y * p
res+= c[n - 1] * y # y or equal to 0 or y > 0
RES = []
if l > 0: # if l not equal to 0 then we can take one more (2 ** (n-1)) and recieve to answer
    RES.append(res + c[n - 1])


for i in range(n - 2, -1, -1):
   p//= 2
   y = l // p
   if y == 1:
       res+= c[i]
       l-= (1 << i)
       if l > 0:
           RES.append(res + c[i])
   else: # y == 0
       if l > 0:
           RES.append(res + c[i])
RES.append(res)
print(min(RES))


    




