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 21:21:36 2018

@author: rohit
"""
import math
def getCost(L):
    global n,b,c
    if L==0:
        return 0
    p = math.floor(math.log(L,2))+1
    ind = b.index(min(b[:p]))
    fill = L//(2**ind)
    rem = L - fill*(2**ind)
    cp = c[ind]*fill
    if p<n:
        return min(cp+getCost(rem),min(c[p:]))
    return cp+getCost(rem)
    
n,L = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
b = [c[i]/(2**i) for i in range(n)]
ans = getCost(L)
p = math.floor(math.log(L,2))+1 
if p<n:
    ans = min(ans,min(c[p:]))
print(ans)