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.

import sys

MAXN = 31
INF = 2**62 - 1 + 2**62

def get_ints():
    return list(map(int, sys.stdin.readline().strip().split()))

def read():
    ints = get_ints()
    n = ints[0]
    L = ints[1]
    cost = get_ints()
    assert len(cost) == n, 'MUST have the same number of elements'

    return n, L, cost

def tobinary(L, minn):
    assert L > 0, 'only handle L > 0'
    digits = []
    while L > 0:
        digits.append(L % 2)
        L = L // 2
    digits.append(0)
    while len(digits) < minn:
        digits.append(0)

    return digits

def get_each_min_cost(cost, length):
    min_cost = []
    for i in range(length):
        min_c = INF
        max_len = min(i+1, len(cost))
        for j in range(max_len):
            min_c = min( (1 << (i-j)) * cost[j], min_c )
        min_cost.append(min_c)

    return min_cost

def solve(digits, min_cost):
    length = len(digits)
    d = [0] * length
    d[0] = min_cost[0] if digits[0] == 1 else 0
    for t in range(1, length):
        if digits[t] == 1:
            d[t] = min(d[t-1] + min_cost[t], min_cost[t]*2)
        else:
            d[t] = min(d[t-1], min_cost[t])

    return d[-1]

if __name__ == '__main__':
    n, L, cost = read()
    digits = tobinary(L, len(cost)+1)
    min_cost = get_each_min_cost(cost, len(digits))
    print(solve(digits, min_cost))
