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;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		StringTokenizer st = new StringTokenizer(sc.nextLine());
		int N = Integer.parseInt(st.nextToken());
		long L = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(sc.nextLine());
		long[] C = new long[31];
		for (int i = 0; i < N; i++) {
			C[i] = Integer.parseInt(st.nextToken());
		}
		for (int i = N; i < C.length; i++) {
			C[i] = C[i - 1] << 1;
		}
		
		for (int i = 1; i < C.length; i++) {
			C[i] = Math.min(C[i], C[i - 1] << 1);
		}
		
		long ans = 0;
		for (int i = 0; i < C.length; i++) {
			if (ans > C[i]) {
				L += 1 << i;
				ans = 0;
			}
			if ((L & (1 << i)) != 0) {
				ans += C[i];
			}
		}
		
		System.out.println(ans);
		
		sc.close();
	}
}