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.
def process(n, c) : #tạo mảng c là min cost để mua đúng 2^i lit (i: 0 -> 30)
    for i in range (n, 30 + 1) : c.append(0)
    for i in range (1, 30 + 1) : 
        if c[i] == 0 : c[i] = c[i - 1] * 2
        else : c[i] = min(c[i], c[i - 1] * 2)
    mn = c[30]
    for i in range (30, 0 - 1, -1) : #trường hợp c[i] < c[i - x]
        c[i] = min(c[i], mn)
        mn   = min(c[i], mn)
    return c

def solve(n, l, c) :
    c = process(n, c)
    case1, case2 = 0, 0

    for i in range(0, 30 + 1) : 
        if (l & (1 << i)) != 0 : case1 += c[i]
    
    case2 = case1; res = case1
    for i in range(0, 30 + 1) :
        if (l & (1 << i)) != 0 : 
            case2 -= c[i]
            res = min(res, case2 + c[i + 1]) 
    return res

def dp009_party_lemonande(n, l, c) : 
    return solve(n, l, c)

if __name__ == '__main__' : 
    n, l = list(map(int, input().split()))
    c    = list(map(int, input().split()))
    print(dp009_party_lemonande(n, l, c))