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 Lemonade {
    public static void main(String[] args) {
        int n,i,j;
        Long l,res=0L,req,ans=4000000000000000000L,extra=0L;

        Scanner scan = new Scanner(System.in);
        n = scan.nextInt();
        l = scan.nextLong();
        Long arr[] = new Long[n+1];

        for(i=0;i<n;i++){
            arr[i] = scan.nextLong();
        }
        for(i=1;i<n;i++){
            arr[i] = mini(arr[i],2*arr[i-1]);
        }
        for(i=n-1;i>=0;i--){
            req = l/(1<<i);
            res += req*arr[i];
            l = l-(req<<i);
            if(l>0) extra = arr[i];
            else extra = 0L;
  //          System.out.println("res="+res+" extra="+extra+" l="+l);
            ans = mini(ans,res+extra);
  //          System.out.println(ans);
        }
        System.out.println(ans);
    }
    static Long mini(Long a,Long b){
        return (a<b)?a:b;
    }

}
