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, li = map(int, input().split())
c = list(map(int, input().split()))

for i in range(n-1):
	if c[i] * 2 < c[i+1]:
		c[i+1] = c[i] * 2

for i in range(n-2, -1, -1):
	if c[i] > c[i+1]:
		c[i] = c[i+1]
'''
p = c[0] * li
for i in range(n):
	if 2**i >= li and p > c[i]:
		p = c[i]
'''

i = n - 1
ans = 0
a = []
while li > 0:
	ans += c[i] * (li >> i)
	li %= 2**i
	if li < 2**i:
		a.append(ans + c[i])
	i-=1

a.append(ans)
print(min(a))

