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.*; 
import java.util.*;
import java.io.*; 
public class MM {
	static Scanner scanner=new Scanner(System.in);
	public static void main(String[] args) {
		int n=scanner.nextInt();
		int l=scanner.nextInt();
		PriorityQueue<Node>pq=new PriorityQueue<Node>((o1,o2)->o1.s>o2.s?1:-1);//按照花费的大小排序
		long ans=Long.MAX_VALUE;
		for(int i=1;i<=n;i++) pq.add(new Node(i, scanner.nextLong()));
		
		long an=0;
		while(!pq.isEmpty()) {
			Node node=pq.poll();
			int x=(int)(l/Math.pow(2, node.i-1));
			an+=x*node.num;
			long s=an;
			l-=(x*Math.pow(2, node.i-1));
			if(l!=0)s+=node.num;
			//System.out.println(s+" "+x);
			ans=Math.min(ans, s);
			if(l==0)break;
		}
		System.out.println(ans);
	}
	static class Node{
		int i;
		long num;
		double s;
		public Node(int i, long num) {
			this.i = i;
			this.num = num;
			s=num/Math.pow(2, i-1);
		}
	}
}












 	  	    	 	  		  					  	 			