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.
def dec(costs,V):
    r=0
    
    price, typ=costs[len(costs)-1]

    fills=int((V-V%int((2**(typ-1))))/int(2**(typ-1)))

    
    r=r+price*fills
    V=V-int(2**(typ-1))*fills
    
    if V<=0:
        return r
    else:#have to buy more
        if len(costs)>1:
            smallerCosts=dec(costs[:len(costs)-1],V)
            if smallerCosts+r<price+r:
                return (smallerCosts+r)
            else:
                return price+r
        else:
            return price+r

trap=input()
t=trap.split(sep=" ")        
n=int(int(t[0]))
V=int(int(t[1]))

tra=input()
t=tra.split(sep=" ")

costs=[]
for k in range(0,n):
    
    costs.append([int(t[k]),k+1])#i =k+1
k=1
while k<len(costs):
    
    
    ivaluec=(costs[k][0]/(2**(costs[k][1]-costs[k-1][1])))
    ivaluep=(costs[k-1][0])
    if ivaluec>ivaluep:
        costs.pop(k)
        k = k-1
    
    k=k+1
k=1
while k<len(costs):
    
    
    if costs[k][0]<costs[k-1][0]:
        costs.pop(k-1)
        k = k-1
    
    k=k+1

r=0

print(dec(costs,V))