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, lit = map(int, input().split())
cost = list(map(int, input().split())) + ([1 << 100] * 33)
n = len(cost)
rlit = lit

for i in range(n):
    for j in range(i-1, -1, -1):
        cost[i] = min(cost[i], cost[j] * (1 << (i - j)))

res = 0

for i in range(n):
    if (1 << i) & lit:
        res += cost[i]

b = []

while lit:
    b.append(lit % 2)
    lit //= 2

rres = res

for i in range(len(b)-1, -1, -1):
    if b[i] == 1:
        for j in range(i-1, -1, -1):
            if b[j] == 0:
                add = cost[j]
                sub = 0
                for k in range(j-1, -1, -1):
                    if b[k] == 1:
                        sub += cost[k]
                res = min(res, rres+add-sub)

for i in range(n):
    if (1 << i) > rlit and cost[i] < res:
        res = cost[i]

print(res)
