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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author hsy
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            long L = in.nextInt();
            int index = 0;
            int[] c = new int[n + 64];
            int[] bits = new int[65];
            long[] dp = new long[n + 64];//dp[i]表示第i个二进制位所需要花费的最小值
            for (int i = 0; i < n; i++) {
                c[i] = in.nextInt();
            }
            /**
             * 将L分解成二进制
             */
            while (L != 0) {
                bits[index++] = (int) L % 2;
                L /= 2;
            }
            /**
             * 求出每个二进制位所需要花费的最小值
             */
            dp[0] = c[0];
            for (int i = 1; i < n || i <= index; i++) {
                if (i < n) {
                    dp[i] = Math.min(dp[i - 1] * 2, c[i]);
                    for (int j = 0; j < i; j++) {
                        dp[j] = Math.min(dp[j], dp[i]);
                    }
                } else {
                    dp[i] = 2 * dp[i - 1];
                }
            }
            for (int i = 0; i <= index; i++) {
                long ans = 0;
                for (int j = 0; j < i; j++) {
                    ans += (dp[j] * bits[j]);
                }
                if (ans > dp[i]) {//如果1--i-1的花费大于第i位的花费则直接购买第i位
                    for (int j = 0; j < i; j++) {
                        bits[j] = 0;
                    }
                    bits[i]++;
                }
            }
            long ans = 0;
            for (int i = 0; i <= index; i++) {
                ans += (dp[i] * bits[i]);
            }
            out.println(ans);
        }

    }
}

