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.
from sys import stdin,stdout
from math import floor,ceil
#stdin.readline()
#int(stdin.readline())
#stdout.write()
#raw_input()
#map(int,raw_input().split())
#map(int,stdin.readline().split())
def f(x,y):
    if x[1]>y[1]:
        return 1
    if x[1]<y[1]:
        return -1
    return 0
n,l=map(int,stdin.readline().split())
a=map(float,stdin.readline().split())
for i in xrange(n):
    t=pow(2,i)
    a[i]=(t,a[i]/t)
a.sort(cmp=f)
t=0
z=0
y=[]
l=float(l)
while t<n and l>=0.0:
    q=floor(l/a[t][0])
    z+=int(q)*int(a[t][1]*a[t][0])
    l-=int(q)*a[t][0]
    y.append(z+int(a[t][1]*a[t][0]))
    t+=1
y.append(z)
y.sort()
print int(y[0])