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.
import math
import bisect
import sys
from collections import OrderedDict
input = sys.stdin.readline
def inp():
	return(int(input()))
def inlt():
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(s[:len(s)-1])
def invr():
	return(map(int,input().split()))
# N, M = inlt()
# A = inlt()
# T = inp()
# MOD = 1000000007
# print(MOD)
# for t in range(T):
N, L = invr()
C = inlt()
for c in range(N-1):
	if C[c]*2 <= C[c+1]:
		C[c+1] = C[c]*2 
out = 10**18
summ = 0
for i in range(N-1,-1,-1):
	a = 2**i
	curr= L//(a)
	L = L%a
	summ += curr*C[i] 
	if L == 0:
		out = min(out,summ)
	else:
		out = min(out,summ+C[i])
print(out)