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.
#!/usr/bin/env python3

def main():
    n, needed = map(int, input().split())
    costs = list(map(int, input().split()))

    done = False
    while not done:
        done = True
        cur = costs[0]
        for i, x in enumerate(costs[1:], 1):
            cur <<= 1
            if cur < x:
                costs[i] = cur
                done = False
            else:
                cur = x
        for i in range(n - 2, -1, -1):
            if costs[i] > costs[i + 1]:
                costs[i] = costs[i + 1]
                done = False

    def calc(total):
        result = 0
        for i in range(31):
            if total & 1 << i:
                if i < n:
                    result += costs[i]
                else:
                    result += costs[-1] << (i - n + 1)
        return result

    result = calc(needed)
    for i in range(30, -1, -1):
        x = (needed >> i | 0x1) << i
        if x >= needed:
            result = min(result, calc(x))

    print(result)

try:
    while True:
        main()
except EOFError:
    pass
