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.ArrayList;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int types = sc.nextInt();
        long liters = sc.nextLong();

        long[] costs = new long[types];
        for(int i = 0; i < types; i++) {
            costs[i] = sc.nextInt();
        }

        ArrayList<Integer> bestSizes = new ArrayList<>();
        bestSizes.add(0);
        double bestCostEach = costs[0];
        for(int i = 1; i < types; i++) {
            double newCostEach = ((double) costs[i]) / (1 << i);
            if(newCostEach < bestCostEach) {
                bestSizes.add(i);
                bestCostEach = newCostEach;
            }
        }

        long currLiters = liters;
        long currCost = 0;
        long bestCost = costs[0] * liters;
        for(int i = bestSizes.size() - 1; i >= 0; i--) {
            int currIndex = bestSizes.get(i);
            int currSize = 1 << currIndex;
            //System.out.println("Index: " + currIndex);
            //System.out.println("Current size: " + currSize);
            //System.out.println("Current cost: " + currCost);
            //System.out.println("Liters remaining: " + currLiters);

            long newCost = currCost;
            long units;
            if(currLiters % currSize == 0) {
                units = currLiters / currSize;
            }
            else {
                units = (currLiters / currSize) + 1;
            }

            newCost += units * costs[currIndex];
            //System.out.println("Total cost: " + newCost);
            if(newCost < bestCost) {
                bestCost = newCost;
            }

            if(currLiters % currSize == 0) {
                currLiters = 0;
                currCost = newCost;
            }
            else {
                currCost += (currLiters / currSize) * costs[currIndex];
                currLiters = currLiters % currSize;
            }
        }

        System.out.println(bestCost);
    }
}
