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 sys
import heapq
import bisect

INF = 10**32
OFFLINE = 0
##sys.setrecursionlimit(INF)

##
if OFFLINE:
	sys.stdin = open("fin.txt", "r")
	sys.stdout = open("fout.txt", "w")
##

def getbit(L):
	bit = []
	for i in range(32):
		bit.append(L&1)
		L = L >> 1
	return bit


#main


n, L = map(int, sys.stdin.readline().split())

a = [INF for i in range(32)]
_a = [0 for i in range(n)]
_a = map(int, sys.stdin.readline().split())
for i in range(n):
	a[i] = _a[i]*1L
dp = [0 for i in range(32)]
dp[0] = a[0]

for i in range(1,32):
	dp[i] = min(a[i], 2*dp[i-1])

A = INF
bit = getbit(L)

bitset = [bit]
for i in range(32):
	if bit[i] == 1:
		continue
	
	_bit = [0]*i
	_bit.append(1)
	_bit.extend(bit[i+1:])
	bitset.append(_bit)

#print bitset

for bit in bitset:
	ans = 0
	for i in range(32):
		ans += dp[i]*bit[i]
	A = min(ans, A)

print A






