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 tt(l,r):
    h = len(l)
    o = l[h-r:]
    c = h-r
    
    for i in range(h-r-1,-1,-1):
        if l[i] == '1':
            o='0'+o
            break
        else:o='1'+o
        c-=1
    else:return'0'
    return l[:c-1]+o
n,l = map(int,input().split())
a = list(map(int,input().split()))
for i in range(1,n):
    a[i]= min(a[i-1]*2,a[i])
r = 2**(n-1)
t = (l//r)*a[-1]
#print(a,l//r)
l %= r
#print(l)
t2 = 0
b = [0]*n
'''
for i in range(n-2,-1,-1):
    if a[i]>=a[i+1]:
        a[i] = a[i+1]
        b[i] = True
'''
l = bin(l)[2:]
t3 = float('inf')

t2 = 0
for i in range(len(l)):
    if l[(-1)-i] == '1':
        t2+=a[i]
t3 = t2
t4  = 0
l = ('0'*(n-len(l)))+l
for i in range(len(l)):
    if l[(-1)-i] == '1':
        t4+=a[i]
    else:
        t3 = min(t2-t4+a[i],t3)
'''
while True:
    t2 = 0
    f = 1
    for i in range(len(l)):
        if l[(-1)-i] == '1':
            t2+=a[i]
        elif f:
            t2 = min(t2,a[i])
            print(l)
            l2 = tt(l,len(l)-i)
            print(l2)
            f = 0
    else:
        print(t2)
        break
    l = l2[:]
    if l =='0':
        break
'''
print(t+t3)
