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.*;

public class Main{
	static long arr[];
	static int val[];
	static long pow[] = new long[40];
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long l = sc.nextInt();
		val = new int[n+1];
		arr = new long[n+1];
		for(int i =0; i <n; i++){
			arr[i] = sc.nextLong();
		}
		for(int i = 0;i<n;i++){
			long ans = arr[i], temp = 2;
			int x = i;
			val[i] = x;
			for(int j = i+1;j<n;j++){
				if(ans*temp>arr[j]) break;
				else{
					val[j] = x;
					i++;
				}
				temp*=2;
			}
		}
		//for(int i = 0;i<n;i++) System.out.print(val[i]+" ");
		//	System.out.println();
		pow[0] = 1;
		for(int i = 1;i<=39;i++) pow[i] = pow[i-1]*2;
		System.out.println(recur(l,n-1));
	}
	static long recur(long l, int x){
		long ans = Long.MAX_VALUE;
		if(x<0) return 0;
		int y = x;
		x = val[x];
		ans = Math.min(ans,((l+pow[x]-1)/pow[x])*arr[x]);
		if(l%pow[x]==0) return ans;
		ans = Math.min(ans, (l/pow[x])*arr[x]+recur(l%pow[x],y-1));
		return ans;
	}
}