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.
#!/usr/bin/python3
# Copyright (C) 2017 Sayutin Dmitry.
#
# This program is free software; you can redistribute it and/or
# modify it under the terms of the GNU General Public License as
# published by the Free Software Foundation; version 3
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with this program; If not, see <http://www.gnu.org/licenses/>.

def solve(a, l):
    if l == 0:
        return 0

    if l == 1:
        return a[0]
    
    k = 0
    while (2 ** k) < l:
        k += 1
    
    return min(a[k], a[k - 1] + solve(a, l - (2 ** (k - 1))))
    

def main():
    n, l = map(int, input().split())
    a = list(map(int, input().split()))

    for i in range(n - 2, -1, -1):
        if a[i] > a[i + 1]:
            a[i] = a[i + 1]
    
    for i in range(1, n):
        if a[i] > 2 * a[i - 1]:
            a[i] = 2 * a[i - 1]

    while len(a) < 35:
        a.append(2 * a[len(a) - 1])

    #print(a)

    print(solve(a, l))
            
main()
