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.*;
public class Main
{
	public static void main(String[] args) {
	    Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		long l=sc.nextLong();
		long[] arr=new long[n];
		for(int i=0;i<n;i++)
		arr[i]=sc.nextLong();
		for(int i=0;i<n-1;i++)
		arr[i+1]=Math.min(arr[i+1],2*arr[i]);
		long ans=Long.MAX_VALUE;
		long sum=0;
		for(int i=n-1;i>=0;i--)
		{
		    long k=l/(1<<i);
		    sum+=k*arr[i];
		    l-=k<<i;
		    if(l!=0)
		    {
		        ans=Math.min(ans,sum+arr[i]);   
		    }
		    else
		    {
		        ans=Math.min(ans,sum);
		        break;
		    }
		}
		System.out.println(ans);
	}
}
