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.Arrays;
import java.util.Scanner;

public class P913C
{
	static long[] cost;
	static long[][] dp;
	static int min;
	static long oo = Long.MAX_VALUE;
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		min = scan.nextInt();
		cost = new long[31];
		Arrays.fill(cost, oo);
		for (int i = 0; i < n; i++)
			cost[i] = scan.nextInt();
		long best = cost[0];
		for (int i = 1; i < cost.length; i++)
		{
			best *= 2;
			if (best <= cost[i])
				cost[i] = best;
			else
			{
				best = cost[i];
				for (int j = 0; j < i; j++)
					cost[j] = Math.min(cost[j], best);
			}
		}
		dp = new long[31][2];
		for (long[] a : dp)
			Arrays.fill(a, -1);
		System.out.println(go(30, 1));
	}
	private static long go(int pow, int bounded)
	{
		if (pow < 0)
			return 0;
		if (dp[pow][bounded] >= 0)
			return dp[pow][bounded];
		long ans = oo;
		int s = bounded == 1 ? ((min & (1 << pow)) != 0 ? 1 : 0) : 0;
		for (int i = s; i <= 1; i++)
		{
			long c = i == 1 ? cost[pow] : 0;
			ans = Math.min(ans, c + go(pow-1, bounded == 0 || i != s ? 0 : 1));
		}
		return dp[pow][bounded] = ans;
	}
}
