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>
using namespace std;
long long a[32] = {0LL};
int main() {
  int n;
  unsigned int L;
  cin >> n >> L;
  int ci;
  for (int i = 0; i < n; i++) {
    cin >> ci;
    a[i] = (long long)ci;
  }
  for (int i = 1; i < 32; i++) {
    if ((i < n) && (a[i] > 2 * a[i - 1])) {
      a[i] = 2 * a[i - 1];
    }
    if (i >= n) {
      a[i] = 2 * a[i - 1];
    }
  }
  long long out = 0LL;
  for (int i = 0; i < 32; i++) {
    if ((L & (1 << i)) > 0) {
      out += (long long)a[i];
    }
  }
  unsigned int t, b;
  long long tl = 0LL;
  bool c = false;
  for (int i = 0; i < 32; i++) {
    t = (1 << 31) - (1 << i);
    b = (L & t) | (1 << i);
    tl = 0LL;
    c = false;
    for (int j = 0; j < 32; j++) {
      if ((b & (1 << j)) > 0) {
        tl += (long long)a[j];
      }
    }
    if (tl < out && b > L) {
      out = tl;
    }
  }
  cout << out << "\n";
  return 0;
}
