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.
#include <bits/stdc++.h>
int take() {
  int n;
  scanf("%d", &n);
  return n;
}
double ttake() {
  double n;
  scanf("%lf", &n);
  return n;
}
long long takes() {
  long long n;
  scanf("%lld", &n);
  return n;
}
int cas;
using namespace std;
bool approximatelyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool essentiallyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) > fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyGreaterThan(float a, float b, float epsilon) {
  return (a - b) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyLessThan(float a, float b, float epsilon) {
  return (b - a) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
int main() {
  int n = take(), L = take();
  long long ans = 1000000000000000000, sum = 0, pw = 1;
  int a[n + 1];
  for (int i = 1; i <= n; i++) a[i] = take();
  for (int i = (2); i <= (n); i += (1)) a[i] = min(a[i], a[i - 1] * 2), pw *= 2;
  for (int i = n; i >= 1; i--) {
    int nd = (L / pw);
    sum += (a[i] * (long long)nd);
    L -= (nd * pw);
    ans = min(ans, sum + (L > 0) * a[i]);
    pw /= 2;
  }
  cout << ans << "\n";
  { return 0; };
}
