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.StringTokenizer;
public class Main{
    public static void main(String[] args) throws IOException{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer(br.readLine());
    	int n = Integer.parseInt(st.nextToken());
    	int L = Integer.parseInt(st.nextToken()), save=0;
    	long sum = 0;
    	int[] cost = new int[n];
    	int[] min = new int[n];
    	st = new StringTokenizer(br.readLine());
    	for(int i=0 ; i<n ; i++)
    		cost[i] = Integer.parseInt(st.nextToken());
    	min[0] = cost[0];
    	for(int i=1 ; i<n ; i++)
    		min[i] = (2*min[i-1]<cost[i]) ? 2*min[i-1] : cost[i];
    	for(int i=n-1 ; i>0 ; i--)
    		if(min[i]<min[i-1])
    			min[i-1] = min[i];
    	for(int i=0 ; i<n ; i++) {
			sum = (min[i] < sum) ? min[i] : sum;
    		if(L == 0)
    			break;
    		if(i < n-1) {
    			sum += (L%2)*(long)min[i];
    			L /= 2;
    			continue;
    		}
    		sum += L*(long)min[i];
    	}
    	System.out.println(sum);
    }
}