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,input().split())
a = list(map(int,input().split()))
#print(0)
b = [0] * n
b[0] = a[0]
if n == 1 and l == 1:
    print(a[0])
else:
    for i in range(1,n):
        b[i] = min(a[i],b[i - 1] * 2)
        if b[i] < b[i - 1]:
            for j in range(i):
                b[j] = min(b[i],b[j])
    c=[0] * n
    c[0] = 1
    s = 0
    z = 100000000000000000000
    for i in range(1,n):
        c[i] = c[i - 1] * 2
    while l > c[n - 1]:
        s += b[n - 1] * (l // c[n - 1])
        l = l % c[n - 1]
        
    #print(0)
    while l > 0:
        for i in range(1,n):
            if l <= c[i] and l >= c[i - 1] :
                z = min(z, s + b[i])
                l -= c[i - 1]
                s += b[i - 1]
                
            #print(l)
    #print(b)
    print(min(z,s))
