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Points: 3
Time limit: 30.0s
Memory limit: 250M

Problem type
Allowed languages
Python

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 2^{i} - 1 liters and costs c_{i} 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 ≤ 10^9) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c_1, c_2, ..., c_{n} (1 ≤ c_{i} ≤ 10^9) — 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.


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