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.
# -*- coding: utf-8 -*-
"""
Created on Mon Jan  8 09:32:18 2018

@author: yanni
"""

n, L = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
size = [2**p for p in range(n)]
for index in range(n):
    if (index > 0):
        c[index] = min(c[index], c[index-1]*2)
for index in range(n):
    if (index > 0):
        temp = n-1-index
        c[temp] = min(c[temp], c[temp+1])
poss = []
base = 0
maxindex = 0
while (size[maxindex] < L and maxindex < n-1):
    maxindex += 1
if (size[maxindex] < L):
    base = (L//size[maxindex]) * c[maxindex]
    L = L % size[maxindex]
if (L == 0):
    print(base)
else:
    poss.append(base + c[maxindex])
    curr = base
    while (L > 0):
        while (size[maxindex] >= 2*L):
            maxindex -= 1
        poss.append(curr + c[maxindex])
        if (maxindex == 0):
            curr += c[0]
            L = 0
        else:
            L -= size[maxindex-1]
            curr += c[maxindex-1]
    poss.append(curr)
    print(min(poss))
    
        
