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.

/**
 * @author: Mehul Raheja
 */
import java.util.*;
import java.io.*;
import static java.lang.Math.*;

public class PartyLemonade {

    /*
        Runtime = O()
     */
    static int N, M, K;
    static String s;
    static StringTokenizer st;
    static int[] d;

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int N = Integer.parseInt(st.nextToken());
        int L = Integer.parseInt(st.nextToken());

        
        long[] c = new long[31];
        Arrays.fill(c, Long.MAX_VALUE);
        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            c[i] = Integer.parseInt(st.nextToken());
        }
        for (int i = 1; i < 31; i++) {
            
            c[i] = min(c[i - 1] * 2, c[i]);
        }
        //System.out.println(Arrays.toString(c));

        long cost1 = 0;
        long temp1 = L;
        for (int k = 0; temp1 > 0; temp1 >>= 1, k++) {
            cost1 += ((temp1 & 1) == 1) ? c[k] : 0;
            //System.out.println(Integer.toBinaryString(temp1) + " " + cost1);
        }

        //System.out.println((1<<31));
        long max = 1<<30;
        max *= 2;
        for (long i = 1, j = 0; i < max; i <<= 1, j++) {
            if (L != (L | i)) {
                //System.out.println("INI1 = " + Integer.toBinaryString(L|i) + " " + j);
                long temp = ((L | i) >> j) << j;
                long cost = 0;
                //System.out.println("INI2 = " + Long.toBinaryString(temp));
                for (int k = 0; temp > 0; temp >>= 1, k++) {

                    cost += ((temp & 1) == 1) ? c[k] : 0;
                    //System.out.println(Integer.toBinaryString(temp) + " " + cost);
                }

                cost1 = min(cost1,cost);
            }
        }
        
        System.out.println(cost1);

    }
}
