# Problem:

time limit per test: 1 second
memory limit per test: 256 megabytes
input: standard input
output: standard output

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 ≤ 301 ≤ 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.

# Notes:

1. Note that for bottle type i, if its cost is larger than twice the cost of bottle type i – 1, we could always buy 2 bottles of type i – 1 instead of 1 bottle of type i. To achieve this, we replace the cost of bottle type i with twice the price of bottle type i – 1 in our cost array c[].
2. Because of last point, we never need to but more than 1 bottle of  type i. Because we can always buy 1 bottle of type i + 1 and it is guaranteed that the cost will never be higher than twice the cost of type i.
3. For the previous two reasons, for any L amount of lemonade we need to purchase, we could always have enough lemonade by buying bottle i if the i th bit of L is one. We also need to buy more than one bottle of the n th bottle if the highest non-zero bit of L is beyond n th bit.
4. Sometimes it is less costly for us to buy more than L amount of lemonade. Let this amount of lemonade be M. There exist a j th bit, for which M’s j th bit is 1 and L’s is 0. Every bit higher than j th is the same in both M and J. Thus, every bit lower than j th in M is zero. Otherwise, we are just buying extra lemonade for no reason.
5. In the implementation, searching for this j th bit only need O(n). Line 22 strips off the highest i bits. Line 25 considers the case which we buys M amount of lemonade. The (L > 0 ) evaluates to 1 when we have non-zero bits lower than i th bit.  Tourist’s implementation is so awesome.