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.
// 05-Mar-2021
import java.util.*;
import java.io.*;

public class PartyLemonade {
	static class FastReader {

		BufferedReader br;
		StringTokenizer st;

		private FastReader() {
			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());
		}

		int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		int[] nextIntArrayOne(int n) {
			int[] a = new int[n + 1];
			for (int i = 1; i < n + 1; i++)
				a[i] = nextInt();
			return a;
		}

		long[] nextLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		long[] nextLongArrayOne(int n) {
			long[] a = new long[n + 1];
			for (int i = 1; i < n + 1; i++)
				a[i] = nextLong();
			return a;
		}

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

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static void main(String[] args) {
		FastReader s = new FastReader();
		StringBuilder str = new StringBuilder();

		int n = s.nextInt();
		long L = s.nextLong();
		long arr[] = s.nextLongArray(n);
		
		for(int i = 0; i < n - 1; i++) {
			arr[i + 1] = Math.min(arr[i + 1], 2 * arr[i]);
		}
		
		long ans = Long.MAX_VALUE;
		long sum = 0;
		
		for(int i = n - 1; i >= 0; i--) {
			long d = L / (1l << i);
			long need = d * arr[i];
			sum += need;
			L -= d * (1l << i);
			long comp = L > 0 ? arr[i] : 0;
			
			ans = Math.min(ans, sum + comp);
			
		}
		
		
		
		
		
		System.out.println(ans);
	}

}
