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,raw_input().split(" "))
cs  =  map(int,raw_input().split(" "))
#cpl = [cs[i]/(2**i) for i in range(n)]

best = []
for i in range(n):
    v1,v2   = cs[i],cs[i]
    for j in range(i+1):
        v1  = min(cs[j]*2**(i-j),v1)
    for j in range(i,n):
        v2  = min(cs[j],v2)
    best.append(min(v1,v2))

ans = 0
ret = []
for i in range(n):
    if 2**i > L:
        ret.append(best[i])
        break
#print best
i = n-1
while L >0:
#    print L,i,ans
    if L >= 2**i:
        if i+1<n:
            ret.append(ans+best[i+1])
        ans += best[i]*int(L/2**i)
        L   %= 2**i
    else:
        i-=1
ret.append(ans)
print min(ret)

		   	 			 	 	  		 				 			  		