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 com.sun.rowset.internal.Row;

import java.io.*;
import java.util.*;


public class Main {
    public static void main(String[] args)  {
       __SolutionBase begin = new __SolutionBase();
       begin.Solve();
    }
}
class __SolutionBase {

    public void Solve() {
        init();
        Solution();
        out.close();
    }
    private void Solution(){

        int n = readInt();
        long L = readLong();
        long[] bottle = new long[n];
        for (int i = 0;i < n;i++) bottle[i] = readLong();
        for (int i = 0;i < n - 1;i++) bottle[i+1] = Math.min(bottle[i]*2,bottle[i+1]);
        long sum = 0;
        long answer = Long.MAX_VALUE;
        for (int i = n - 1; i > -1 ;i--){
            long shouldBuy = L/(1 << i);
            L -= shouldBuy << i;
            sum += shouldBuy*bottle[i];
            answer = Math.min(answer, sum + (L > 0? 1 : 0)*bottle[i]);
        }
        out.println(answer);
    }

    /**
     * readLong, readDouble and other are similar
     *
     * @return int type from input
     */
    private int readInt() {
        return Integer.parseInt(readString());
    }

    private long readLong() {
        return Long.parseLong(readString());
    }
    /**
     * Tokens are separated by space or endExclusive of line
     *
     * @return non-empty string token from input (or null if there is no any token)
     */
    private String readString() {
        while (!tok.hasMoreTokens()) {
            String nextLine = readLine();
            if (null == nextLine) return null;
            tok = new StringTokenizer(nextLine);
        }

        return tok.nextToken();
    }

    /**
     * @return whole line from input
     */
    private String readLine() {
        try {
            return in.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tok = new StringTokenizer(" ");

    /**
     * For local testing I use pair 'input.txt'/'output.txt' in the project root
     */
    private void init() {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        } catch (FileNotFoundException e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
    }
}

