913_C. Party Lemonade
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.
```python
n = input().split()
l = int(n[1])
n = int(n[0])
c = input().split()
cperl = []
for i in range(n):
    c[i] = int(c[i])
    cperl.append([(c[i])/2**i, 2**i, i])
cperl.sort()
f = False
nl = [0]
nc = [0]
i = 0
while not f:
    p = nl[i]
    nl[i] += cperl[i][1] * ((l-p)//cperl[i][1])
    nc[i] += c[cperl[i][2]] * ((l-p)//cperl[i][1])
    if nl[i] == l:
        f = True
        break
    nl.append(nl[i])
    nc.append(nc[i])
    nl[i] += cperl[i][1]
    nc[i] += c[cperl[i][2]]
    i += 1
    if i == n:
        break
print(min(nc))
```