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.*;
import java.io.*;

public class File {
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }

            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
	
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		
		
		int n = sc.nextInt();
		long L = sc.nextLong();
		
		long[] costs = new long[n];
		for (int i = 0; i < n; i++) {
			costs[i] = sc.nextLong();
		}
		
		
		for (int i = 0; i + 1 < n; i++) {
			costs[i+1] = Math.min(costs[i+1], 2*costs[i]);
		}
		
		
		long answer = Long.MAX_VALUE;
		long total = 0;
		
		for (int i = n-1; i >= 0; i--) {
			long div = L / (1 << i);
			total += (div * costs[i]);
			
			L -= (div * (1 << i));
			
			// Update answer.
			// If there is remaining, we could add another one of this one.
			// Or, it'll be cheaper for smaller bottles.
			answer = Math.min(answer, total + (L > 0 ? 1 : 0) * costs[i]);
		}
		
		
		out.println(answer);
		
		
		
		
		
		
		
		
		
		
		out.close();
	}
}











	
