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.*;

public class scorify {

    
    public static void main(String[] args){
        Scanner  in = new Scanner(System.in);
        int n= in.nextInt();
        long L = in.nextLong();
        long[] Costs = new long[n];
        for(int i=0; i<n; i++){
            Costs[i] = in.nextLong();
        }
        for(int i=1; i<n; i++){
            Costs[i] = Math.min(2L*Costs[i-1] , Costs[i]);
        }
        
        long ans = Long.MAX_VALUE;
        long s = 0L;
        for(int i = n-1; i>-1; i--){
            long f = L/(1<<i);
            s += f*Costs[i];
            L -= f*(1<<i);
            ans = Math.min(ans,s+(L>0?Costs[i]:0L));
        }
        
        System.out.println(ans);
    }
}