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())
c=list(map(int,input().split()))
for i in range(1,n):
  c[i]=min(c[i],c[i-1]*2)
x=0
d=l
for sht in range(n-1):
  if (l>>sht)&1:
    x+=c[sht]
    d-=2**sht
x+=c[-1]*(d//(2**(n-1)))
ans=x
u="0"*(n-len(str(bin(l)[2:])))+str(bin(l)[2:])
uu=len(u)
for j in range(uu): # at j_bit from the top
  if u[j]=="1":
    continue
  m=u[:j]+"1"+"0"*(uu-1-j)
  m=int(m,2)
  mm=m
  x=0
  for sht in range(n-1):
    if (m>>sht)&1:
      x+=c[sht]
      m-=2**sht
  x+=c[-1]*(m//(2**(n-1)))
  x=min(x,c[-1]*((mm-1)//(2**(n-1))+1))
  ans=min(ans,x)
ans=min(ans,c[-1]*((l-1)//(2**(n-1))+1))
print(ans)