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.Scanner;

public class C {

/*
4 12
20 30 70 90

4 3
10000 1000 100 10

4 3
10 100 1000 10000

5 787787787
123456789 234567890 345678901 456789012 987654321

3 100
1 2 3

2 3
2 3

 */
	
	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		int n = reader.nextInt();
		int L = reader.nextInt();
		int[] c = new int[n];
		double[] ratio = new double[n];
		int[] v = new int[n];
		for(int i = 0; i < n; i++) {
			c[i] = reader.nextInt();
			v[i] = (1 << i);
			ratio[i] = (1 << i) / (double)c[i];
		}
		for(int i = 1; i < n; i++) {
			if(2 * c[i - 1] < c[i]) {
				c[i] = 2 * c[i - 1];
			}
		}
		long minCost = Long.MAX_VALUE;
		long cost = 0;
		int i = n - 1;
		while(i >= 0) {
			if(L >= v[i]) {
				long r = L / v[i];
				L -= r * v[i];
				cost += r * c[i];
			} else {
				minCost = Math.min(minCost, cost + (L > 0 ? c[i] : 0));
				i -= 1;
			}
		}
		System.out.println(minCost);
		reader.close();
	}
	
}
