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.*;
public class C {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		long min = Long.MAX_VALUE;
		long cum = 0;
		long l = sc.nextLong();
		long[] values = new long[n];
		for(int i =0; i<n; i++) {
			values[i] = sc.nextLong();
		}
		
		while(n>0) {
			if(l==0)
				break; // nothing remains
			double price = (double)values[n-1]/(Math.pow(2, n-1));
			boolean flag = true;
			for(int i=0; i<n-1; i++) {
				double others = (double)values[i]/(Math.pow(2, i));
				if(others < price) {
					flag = false;
					break;
				}
			}
			if(!flag) {
				n--;
				continue;
			}
			long j = l/(long)(Math.pow(2, n-1));
			if(j*Math.pow(2, n-1) < l)
				j++;
			min = Math.min(cum+j*values[n-1], min);
			cum += (j-1)*values[n-1];
			l -= (j-1)*Math.pow(2, n-1);
			n--;
		}
		
		System.out.println(min);
 	}
}
