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 java.util.*;
import java.io.*;

public class P913C
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = in.nextInt();
        long required = in.nextLong();

        long[] cost = new long[32];
        for(int i = 0; i < n; ++i)
        {
            cost[i] = in.nextLong();
        }

        for(int i = 1; i < n; ++i)
        {
            long t = cost[i - 1] * 2;
            if(t < cost[i])
            {
                cost[i] = t;
            }
        }
        for(int i = n; i < cost.length; ++i)
        {
            cost[i] = cost[i - 1] * 2;
        }

        long ans = Long.MAX_VALUE;
        String b = Long.toBinaryString(required);
        b = new StringBuilder(b).reverse().toString();

        for(int i = b.length(); i < cost.length; ++i)
        {
            if(cost[i] < ans)
            {
                ans = cost[i];
            }
        }
        long prefix = 0;
        for(int i = b.length() - 1; i >= 0; --i)
        {
            if(b.charAt(i) == '0')
            {
                long t = prefix + cost[i];
                if(t < ans)
                {
                    ans = t;
                }
            }
            else
            {
                prefix = prefix + cost[i];
            }
        }
        if(prefix < ans)
        {
            ans = prefix;
        }

        out.println(ans);

        out.close();
    }
}
