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.io.*;
import java.util.*;
import java.math.*;
public class lemonade
{
    public static void main(String args[])throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String s[]=br.readLine().trim().split(" ");
        int n=Integer.parseInt(s[0]);
        int l=Integer.parseInt(s[1]);
        int arr[]=new int[n];
        s=br.readLine().split(" ");
        for(int i=0;i<n;i++)
        {
            arr[i]=Integer.parseInt(s[i]);
        }
        BigInteger min=new BigInteger(Long.toString(Long.MAX_VALUE/2));
        int limit=n,c=0,rem=l;BigInteger ccc=new BigInteger("0");
        while(limit!=0 && rem!=0)
        {
            double costperlit=Long.MAX_VALUE-1;int ind=-1;
            for(int i=0;i<limit;i++)
            {
                double cost=arr[i]/Math.pow(2,i);
                if(cost<costperlit)
                {
                    costperlit=cost;
                    ind=i;
                }
            }
            if(c==0)
            {
                BigInteger nb=new BigInteger(Integer.toString((int)Math.ceil(rem/Math.pow(2,ind))));
                min=nb.multiply(BigInteger.valueOf(arr[ind]));
                c++;
                BigInteger less=new BigInteger(Integer.toString((int)Math.floor(rem/Math.pow(2,ind))));
                ccc=ccc.add(less.multiply(BigInteger.valueOf(arr[ind])));
                int les=(int)Math.floor(rem/Math.pow(2,ind));
                rem=rem-(int)(les*Math.pow(2,ind));
                limit=ind;
            }
            else
            {
                BigInteger nb=new BigInteger(Integer.toString((int)Math.ceil(rem/Math.pow(2,ind))));
                BigInteger temp=ccc.add(nb.multiply(BigInteger.valueOf(arr[ind])));
                min=min.min(temp);
                int less=(int)Math.floor(rem/Math.pow(2,ind));
                ccc=ccc.add(BigInteger.valueOf(less).multiply(BigInteger.valueOf(arr[ind])));
                rem=rem-(int)(less*Math.pow(2,ind));
                limit=ind;
            }
        }
        min=min.min(ccc);
        System.out.println(min);
    }
}