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.
if __name__ == "__main__":
    n, L = [int(a) for a in input().strip().split()]
    costs_n = [int(a) for a in input().strip().split()]

    costs = [0 for i in range(31)]
    for i in range(n):
        costs[i] = costs_n[i]

    for i in range(1, n):
        costs[i] = min(costs[i], costs[i-1] * 2)
    
    for i in range(n-2, -1, -1):
        costs[i] = min(costs[i], costs[i+1])

    for i in range(n, 31):
        costs[i] = costs[i-1] * 2
        
    res = 0
    for i in range(31):
        if res > costs[i]:
            res = costs[i]
        if (L & (2 ** i)):
            res += costs[i]
    
    print(str(res))
 		 	 	   	   		     			 		