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;
int n, dui;
long long L, cnt;
long long Ci[35];
long long ans;
void Dui() {
  long long Bi = 2;
  for (dui = 1; Bi <= L; dui++) Bi <<= 1;
  dui += 1;
}
int main() {
  cin >> n >> L;
  Dui();
  for (int i = 1; i <= n; i++) cin >> Ci[i];
  for (int i = n - 1; i >= 1; i--) {
    if (Ci[i] > Ci[i + 1]) Ci[i] = Ci[i + 1];
  }
  for (int i = 2; i <= n; i++)
    if (Ci[i] > Ci[i - 1] * 2) Ci[i] = Ci[i - 1] * 2;
  for (int i = n + 1; i <= dui; i++) Ci[i] = (Ci[i - 1] << 1);
  while (L) {
    cnt++;
    if (L & 1) {
      ans += Ci[cnt];
    } else if (Ci[cnt] < ans)
      ans = Ci[cnt];
    L >>= 1;
  }
  if (cnt < dui && Ci[cnt + 1] < ans) ans = Ci[cnt + 1];
  cout << ans << endl;
  return 0;
}
