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 solve(a1,b1, cur_sum, cur_L, idx):
    if cur_L == 0:
        return cur_sum
    if idx == len(a1)-1:
        if cur_L > 0:
            ost = cur_L%b1[-1] > 0
            if ost:
                cur_sum += a1[-1]
            cur_sum += a1[-1]* (cur_L//b1[-1])
            return cur_sum
        else:
            return cur_sum
    else:
        ost = cur_L%b1[idx] > 0
        if ost:
            s1 = solve(a1,b1, cur_sum+a1[idx]*((cur_L//b[idx]) + 1), 0,idx+1)
            s2 = solve(a1,b1, cur_sum+a1[idx]*((cur_L//b[idx])), cur_L % b1[idx],idx+1)
            return min(s1,s2)
        else:
            return solve(a1,b1, cur_sum+a1[idx]*((cur_L//b[idx])), cur_L % b1[idx],idx+1)
            

            
n, L = map(int, input().split())
a = list(map(int,input().split()))
b = [2**i for i in range(len(a))]
for i in range(len(a)):
        for j in range(len(a) - 1, i, -1):
            if a[j]/b[j] < a[j-1]/b[j-1]:
                a[j], a[j-1] = a[j-1], a[j]
                b[j], b[j-1] = b[j-1], b[j]
ans = solve(a,b,0,L,0)
print (ans)

        
