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 A7 {
	
	 public static void main(String []args){
	        Scanner sc=new Scanner(System.in);
	        int n =sc.nextInt();
	        long L=sc.nextLong();
	        long[] a=new long[n];
	        for(int i=0;i<n;i++)
	        	a[i]=sc.nextLong();
	        
	        boolean[] leaf = new boolean[n];
	        ArrayList<ArrayList<Integer>> al = new ArrayList<>();
	        for(int i=0;i<n;i++)
	        	al.add(new ArrayList<>());
	       
	        for(int i = 0;i < n-1;i++){
				a[i+1] = Math.min(a[i+1], a[i]*2);
			}
			for(int i = n-2;i >= 0;i--){
				a[i] = Math.min(a[i], a[i+1]);
			}
			long ans = Long.MAX_VALUE;
			
			long cur = 0;
			for(int i = n-1;i >= 0;i--){
				long x = L / (1L<<i);
				//System.out.println(x+" "+ "++++");
				cur += a[i] * x;
				ans = Math.min(ans, cur + a[i]);
				//System.out.println(ans+" "+ "+++" +cur);
				L -= x<<i;
				//System.out.println(L+" "+ "++");
			//	System.out.println(ans+" "+ "+"); 
				
			}
			ans = Math.min(ans, cur);
			System.out.println(ans);
	        }
	        
	                
	            
	     }