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;

/**
 * Created on 08/01/2018.
 *
 * @author ThanhKM thanhkhongminh@gmail.com
 */
public class PC {
    static int n, L;
    static double[] per;
    static long[] c;


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        n = sc.nextInt();
        L = sc.nextInt();

        c = new long[n+1];
        for(int i = 1; i <= n; i++) {
            c[i] = sc.nextInt();
        }

        per = new double[n+1];
        for (int i = 1; i <= n; i++) {
            per[i] = c[i] * 1.0 / (1<<(i-1));
        }

        /*for (int i = 1; i <= n; i++) {
            System.out.print(per[i] + " ");
        }
        System.out.println();*/

        int len = Integer.toBinaryString(L).length();
        long cost = Long.MAX_VALUE;
        for (int i = len+1; i <= n; i++) {
            if (cost > c[i]) {
                cost = c[i];
            }
        }

        System.out.println(Math.min(cost, minCost(L, len)));
    }

    private static long minCost(int volume, int idx) {
        if (volume == 0)
            return 0;

        //int len = Integer.toBinaryString(volume).length();

        int minIdx = 1;
        double minPer = per[minIdx];

        for (int i = 1; i <= idx && i <= n; i++) {
            if (per[i] < minPer) {
                minIdx = i;
                minPer = per[i];
            }
        }

        int take = volume / (1<<(minIdx-1));
        int rest = volume - (take << (minIdx-1));
        //System.out.println(take + " " + minIdx + " " + rest);

        return take * c[minIdx] + Math.min(c[minIdx], minCost(rest, minIdx-1));
    }
}
