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.*;
import java.io.*;
public class C {
    long[] C;
    HashMap<String, Long> map = new HashMap<>();
    long min(long L, int i) {
        if(i == 0) return L*C[0];
        String s = L + ";" + i;
        if(map.containsKey(s)) return map.get(s);
        long p2 = 1L<<i;
        long amount = (L+ p2 - 1)/p2;
        long min = C[i]*amount;
        long a2 = amount - 1;
        long min2 = Math.min(min, min(L-a2*p2, i - 1) + a2*C[i]);
        if (min2 < min) 
            min = Math.min(min2, min(L, i - 1));
        map.put(s, min);
        return min;
    }
    void solve(BufferedReader in) throws Exception {
        long[] xx = toInts(in.readLine());
        int n = (int)xx[0];
        long L = xx[1];
        C = toInts(in.readLine());
        System.out.println(min(L, n-1));

        
    }
    long toInt(String s) {return Integer.parseInt(s);}
    long[] toInts(String s) {
        String[] a = s.split(" ");
        long[] o = new long[a.length];
        for(int i = 0; i<a.length; i++) o[i] = toInt(a[i]);
        return o;
    }
    void e(Object o) {
        System.err.println(o);
    }
    public static void main(String[] args) throws Exception{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        (new C()).solve(in);
    }
}
