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.Scanner;

/**
 * http://codeforces.com/problemset/problem/913/C
 * @author Alan Yan
 *
 */
public class Task913C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();int L = sc.nextInt();
		
		long[] costs = new long[n];
		long ans = Long.MAX_VALUE;
		long sum = 0;
		for(int i = 0; i < n; i++) 
			costs[i] = sc.nextLong();
		
		for(int i = 0; i < n-1; i++) {
			costs[i+1] = Math.min(costs[i+1], 2*costs[i]);
		}
		
		for(int i = n-1; i >= 0; i--) {
			int need = L / (1 << i);
			sum += (long) need * costs[i];
			L -= need << i;
			ans = Math.min(ans, sum + bool(L > 0) * costs[i]);
		}
		
		System.out.print(ans);
		sc.close();
	}
	static int bool(boolean x) {
		if(x) return 1;return 0;
	}
}
