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=[int(i) for i in input().split()]
L=list(bin(L).replace('0b',''))
L.reverse()
c=[int(i) for i in input().split()]
      
l=0
C=[]
for i in range(len(L)):

    c_l={}
    for j in range(n):
        if j>i:
            c_l[c[j]/2**i]=j
        elif j<=i:
            c_l[c[j]/2**j]=j
    c_ls=list(c_l.keys())
    m=c_l[min(c_ls)]
    if m>i:
        Ci=c[m]
    if m<=i:
        Ci=c[m]/(2**m)*(2**i)
    Cc1=l+int(Ci*int(L[i]))
    
    c_l.clear()
    for j in range(n):
        if j>(i+1):
            c_l[c[j]/2**(i+1)]=j
        elif j<=(i+1):
            c_l[c[j]/2**j]=j
    c_ls=list(c_l.keys())
    m=c_l[min(c_ls)]
    if m>(i+1):
        Ci=c[m]
    if m<=(i+1):
        Ci=c[m]/(2**m)*(2**(i+1))
    Cc2=int(Ci*int(L[i]))

    x=min(Cc1,Cc2)
    l=l+x
    if x==0:
        C.append(l)
    else:
        C.append(x)
        l=x

print(int(C[-1]))
