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.Arrays;
import java.util.Scanner;

public class c {
	static int N, L;
	static long[] c;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		N = in.nextInt();
		L = in.nextInt();
		
		c = new long[31];
		Arrays.fill(c, Long.MAX_VALUE);
		for(int n=0;n<N;n++){
			c[n] = in.nextInt();
		}
		
		// let's modify the costs
		// observation 1: bottles of higher volume can be replaced with multiple bottles if cost is lower
		// observation 2: bottles of lower vol can be replaced with bottles of higher vol if lower is too expensive
		
		for(int i=0;i<31;i++){
			for(int n = i+1; n < 31; n++){
				// 1
				long prop = (c[i] << (n-i));
//				System.out.println(i+"->"+n);
				if(prop < c[n]){
					
//					System.out.println(c[n]+"->"+prop);
					c[n] = prop;
//					System.out.println(c[n]);
				}
				
				// 2
				if(c[i] > c[n]){
					c[i] = c[n];
				}
			}
		}
		
//		System.out.println(Arrays.toString(c));
		
		long bareCost = Long.MAX_VALUE;
		
		// i don't even know what this does
		// try to hack me; i triple dog dare you
		for(long l = L; l < 2_000_000_001; l = l + (l & -l)){
			long cost = 0;
			for(int n=0;n<31;n++){
				if((l >> n & 1) == 1){
					cost += c[n];
				}
			}
			bareCost = Math.min(bareCost, cost);
		}
		// this is not necessarily the best way to do it
		
		// if any of the bottles after the high bit are cheaper than the total, then let's use that instead
		long minAfter = Long.MAX_VALUE;
		for(int n = Integer.highestOneBit(L)+1; n < N; n++){
			minAfter = Math.min(minAfter, c[n]);
		}
		
		System.out.println(Math.min(bareCost, minAfter));
	}
	
	
}
