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()); d=32*[10737418240000000000]; m=[10737418240000000000]
c=list(map(int, input().split())); d[0]=c[0]
if n<=30:
    for t in range(n, 32):
        c.append(1073741824000000000)
for t in range(32):
    if (t!=0):
        if c[t]<=2*d[t-1]:
            d[t]=c[t]
        else:
            d[t]=2*d[t-1]
x=len(bin(l))-2
ans=0; cur=0
s=bin(l)[2:]
for t in range(x):
    if s[t]=='1':
        ans+=d[x-int(t)-1]
    else:
        cur=ans+d[x-int(t)-1]
        m.append(cur)
print (min(ans, min(m), min(d[x:])))
    

        
    
    
