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;

public class WorkFile {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), l = s.nextInt();
        long[] array = new long[n];
        array[0] = s.nextInt();
        for (int i=1; i<n; i++) {
            long x = s.nextInt();
            array[i] = Math.min(array[i-1]*2, x);
        }
        long res = 0;
        long x = (long) Math.pow(2, n-1);
        long[] ans = new long[n];
        for (int i=n-1; i>-1; i--) {
            ans[i] = l/x*array[i];
            l%=x;
            x/=2;
        }
        for (int i=1; i<n; i++) {
            ans[i] = Math.min(ans[i]+array[i], ans[i]+ans[i-1]);
        }
        System.out.println(ans[n-1]);
    }
}