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.
a,b=map(int,input().split())
z=list(map(int,input().split()))
import math
pre=[z[0]]
for i in range(1,len(z)):
    pre.append(min(z[i],2*pre[-1]))

    
q=[1]
for i in range(len(z)-1):
    q.append(q[-1]*2)
cost=0
p=len(q)-1
pos=[]

mini=math.inf
while(1):
    if(b>=q[p]):
        nos=b//q[p]
        rem=b%q[p]
        
        cost+=nos*pre[p]
        if(rem==0):
            print(min(mini,cost))
            break;
        else:
            pos.append(cost+pre[p])
            mini=min(mini,min(pos))
            b=rem
            p=p-1
    else:
        pos.append(cost+pre[p])
        p=p-1
        mini=min(mini,min(pos))
        
        

        






