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, raw_input().split())
orig_costs = map(int, raw_input().split())
n = max(n, len(bin(L)[2:]))
costs = [float('inf')] * (n + 1)
for i in xrange(len(orig_costs)):
    costs[i] = orig_costs[i]
cum_sum = 0
for i in xrange(n):
    if 2*costs[i] < costs[i + 1]:
        costs[i + 1] = 2 * costs[i]
    cum_sum += costs[i]
    if cum_sum + costs[0] < costs[i + 1]:
        costs[i + 1] = cum_sum + costs[0]
##print costs[:len(orig_costs)]
#print costs[-5:]
for i in xrange(n-2, -1, -1):
    costs[i] = min(costs[i], costs[i + 1])
bits_order = bin(L)[2:][::-1]
#print bits_order
try_cost = 0
for i, b in enumerate(bits_order):
    if b == '1':
        try_cost += costs[i]
    try_cost = min(try_cost, costs[i + 1])
for i in range(len(bits_order), len(costs)):
    try_cost = min(try_cost, costs[i])
print try_cost
