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.
// package com.company.codeforces;


import java.io.*;
import java.util.*;

public class Solution {



    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n=input.nextInt();
        long l=input.nextLong();
        long c[]=new long[n];
        long indexmincost[]=new long[31];
        for (int i = 0; i <n ; i++) {
            c[i]=input.nextLong();
        }
        indexmincost[0]=c[0];
        for (int i = 1; i <n ; i++) {
            indexmincost[i]=Math.min(indexmincost[i-1]*2,c[i]);
        }
        for (int i = n; i <=30; i++) {
            indexmincost[i]=indexmincost[i-1]*2;

        }
        long dp[]=new long[31];
        for (int i = 0; i <=30; i++) {
            if (i==0) {
                dp[0]=0;
            }else {
                dp[i]=Math.min(dp[i-1],indexmincost[i]);
            }

            long temp=1<<i;
            if ((temp&l)==temp){
                dp[i]+=indexmincost[i];

            }
            
        }
        System.out.println(dp[30]);
    }

}