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.
/**
 * DA-IICT
 * Author : Savaliya Sagar
 */
import java.io.*;
import java.math.*;
import java.util.*;


public class C913
{
	void solve()
	{
		int n = ni();
		int L = ni();
		long l[] = new long[n];
		long s[] = new long[n];
		for(int i=0;i<n;i++) {
			l[i] = ni();
			s[i] = (1<<i);
		}
		for(int i=1;i<n;i++) {
			l[i] = Math.min(l[i-1]*2,l[i]);
		}
		long cost = 0;
		long k = L/s[n-1];
		cost = k*l[n-1];
		L -= k*s[n-1];
		int ll = n/2;
		long left[] = new long[1<<ll];
		for(int i=0;i<(1<<ll);i++) {
			long c = 0;
			for(int j=0;j<ll;j++) {
				if(((i>>j)&1)==1)
					c += l[j];
			}
			left[i] = c;
		}
		
		int rr = n - n/2;
		long right[] = new long[1<<rr];
		for(int i=0;i<(1<<rr);i++) {
			long c = 0;
			for(int j=0;j<rr;j++) {
				if(((i>>j)&1)==1)
					c += l[j+ll];
			}
			right[i] = c;
		}
		for(int i=(1<<rr)-2;i>=0;i--) {
			right[i] = Math.min(right[i+1], right[i]);
		}
		long tc = Long.MAX_VALUE;
		double o = (double)(1<<ll);
		for(int i=0;i<(1<<ll);i++) {
			long p = (long)Math.ceil((L-i)/o);
			if(p<right.length) {
				tc = Math.min(tc,left[i]+right[(int)p]);
			}
		}
		out.println(tc+cost);
	}
	public int ni()
	{
		return Integer.parseInt(next());
	}
	public long nl()
	{
		return Long.parseLong(next());
	}
	private void tr(Object... o)
	{
		if (debug == 1)
			System.out.println(Arrays.deepToString(o));
	}
	
	int debug = 1;
	void run() throws Exception
	{
		out = new PrintWriter(System.out);
		reader = new BufferedReader(new InputStreamReader(System.in), 32768);
		tokenizer = null;
		solve();
		out.flush();
	}
	
	BufferedReader reader;
	StringTokenizer tokenizer;
	PrintWriter out;
	
	public static void main(String[] args) throws Exception
	{
		new C913().run();
	}
	
	public String next()
	{
		while (tokenizer == null || !tokenizer.hasMoreTokens())
		{
			try
			{
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e)
			{
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
}
