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.*;
import java.util.*;

public class partC {
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		int n = (int)in.nextLong();
		long L = in.nextLong();
		
		long[] c = new long[n];
		
		for(int i = 0; i < n; i++) {
			c[i] = in.nextLong();
		}
		
		in.close();
		
		for(int i = 1; i < n; i++) {
			if(c[i] > 2*c[i-1]) {
				c[i] = 2*c[i-1];
			}
		}
		
		long[] p2 = new long[n];
		long counter = 1;
		for(int i = 0; i < n; i++) {
			p2[i] = counter;
			counter *= 2;
		}
		
		
		long min = Long.MAX_VALUE;
		
		long big = 0;
		long cost = 0;
		
		long bulk = (L / p2[n-1]) -1;
		big = p2[n-1] * bulk;
		cost = c[n-1] * bulk;
		
		while(L > big + p2[n-1]) {
			big += p2[n-1];
			cost += c[n-1];
		}
		
		for(int j = n-1; j >= 0; j--) {
			boolean keep;
			if(L - big > p2[j]-1) {
				keep = true;
			}else {
				keep = false;
			}
			
			big += p2[j];
			cost += c[j];
			if((big >= L) && (cost < min)) {
				min = cost;
			}
			
			if(keep == false) {
				big -= p2[j];
				cost -= c[j];
			}
		}
		
		System.out.println(min);
		
		
	}	
	
}
