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.io.*;
import java.math.BigInteger;
import java.util.*;


public class C {

  long solve(int l, int[] a, long[] c, int n) {
    if (n<0) {
      return 0;
    }
    if (a[n]<l) {
      return Math.min(c[n]*2, c[n]+solve(l-a[n], a, c, n-1));
    } else {
      if (a[n]==l)
        return c[n];
      else
        return Math.min(c[n], solve(l, a, c, n-1));
    }
  }

  void solve() throws IOException {
    int n=nextInt();
    int l=nextInt();
    long[] c=new long[n];
    for(int i=0;i<n;i++){
      c[i]=nextInt();
    }
    if (n==1) {
      out.println(c[0]*l);
      return;
    }
    int[] a=new int[n];
    a[0]=1;
    for(int i=1;i<n;i++){
      a[i]=a[i-1]*2;
      if(c[i-1]*2<c[i])c[i]=c[i-1]*2;
    }
    long ans=0;
    while (l>a[n-1]){
      l-=a[n-1];
      ans+=c[n-1];
    }
    ans+=solve(l,a,c,n-1);
    out.println(ans);
  }

  public static void main(String[] args) throws IOException {
    new C().run();
  }

  void run() throws IOException {
    reader = new BufferedReader(new InputStreamReader(System.in));
//		reader = new BufferedReader(new FileReader("input.txt"));
    tokenizer = null;
    out = new PrintWriter(new OutputStreamWriter(System.out));
//		out = new PrintWriter(new FileWriter("output.txt"));
    solve();
    reader.close();
    out.flush();

  }

  BufferedReader reader;
  StringTokenizer tokenizer;
  PrintWriter out;

  int nextInt() throws IOException {
    return Integer.parseInt(nextToken());
  }

  long nextLong() throws IOException {
    return Long.parseLong(nextToken());
  }

  double nextDouble() throws IOException {
    return Double.parseDouble(nextToken());
  }

  String nextToken() throws IOException {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine());
    }
    return tokenizer.nextToken();
  }
}
