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 s = new Scanner(System.in);
		int n= s.nextInt();
		long lt = s.nextLong();
		long arr[] = new long[n+1];
		for(int i = 0;i<n;i++)
			arr[i] = s.nextLong();
		long pow[] = new long[n+1];
		for(int i = 1;i<=n;i++){
			pow[i] = (long)Math.pow(2, i-1);
		}
		long val = Long.MAX_VALUE;
		for(int i = 1;i<n;i++){
			arr[i] = Math.min(arr[i], 2*arr[i-1]);
		}
		int temp;
		long dp = 0;
		long ans =  Long.MAX_VALUE;
		for(int i = n;i>=1;i--){
			dp += (lt/pow[i])*arr[i-1];
			if(lt%pow[i]!=0)
				ans = Math.min(ans, dp+arr[i-1]);
			else
				ans = Math.min(ans, dp);
			lt-=(lt/pow[i])*pow[i];
			
				
		}
		System.out.println(ans);
	}
}