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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

import static java.lang.Math.pow;

public class Main {
    private static boolean [] b;
    public static void main(String[]args) {
        //System.out.println((long)Math.pow(10,9));
        //System.out.println((long)Double.MAX_VALUE);
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        long l = scanner.nextInt();
        long c [] = new long[n];
        long p [] = new long[n];

        for(int i=0;i<n;i++){
            c[i] = scanner.nextLong();
            p[i] = (long) pow(2,i);
            if(i>0){
                c[i] = Math.min(c[i], 2 * c[i-1]);
            }
        }

        //for(int i=0;i<n;i++){
            //System.out.println(p[i]+ " " + c[i]);
        //}
        //System.out.println();
        ArrayList<Long> ress = new ArrayList<>();
        long res = 0;
        for(int i = n - 1; i >= 0; i--){
            long k = l/p[i];
            ress.add(l % p[i] != 0 ? res+(k+1)*c[i] : res + k * c[i]);

            if(l>=p[i]){
                res+=k*c[i];
                l-=k*p[i];
            }
        }
        //System.out.println(ress);
        System.out.println(Collections.min(ress));
    }
}