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 random
import functools
import re
import itertools
import math
import copy
import bisect
import Queue
from collections import Counter

sys.setrecursionlimit(1000000)

def debug(*_):
    for x in _:
        print >> sys.stderr, x,
    print >> sys.stderr


S = {}

def solve(l, costs):
    if len(costs) == 1:
        return costs[0] * l
    
    if l == 0:
        return 0
    
    if (l, len(costs)) in S:
        return S[l, len(costs)]
    
    res = min(
        costs[0] * (l % 2) + solve(l / 2, costs[1:]),
        solve(l / 2 + 1, costs[1:])
    )
    S[l, len(costs)] = res
    return res


if __name__ == "__main__":
    n, l = [int(_) for _ in sys.stdin.readline().split()]
    costs = [int(_) for _ in sys.stdin.readline().split()]
    
    for i in xrange(1, n):
        costs[i] = min(costs[i], 2*costs[i-1])
    
    print solve(l, costs)
    
