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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long l;
  cin >> l;
  long long cost[31];
  for (int i = 0; i < n; i++) cin >> cost[i];
  for (int i = 1; i < n; i++) cost[i] = min(cost[i], 2 * cost[i - 1]);
  long long tcost = 0, ans = LLONG_MAX;
  int last = 0;
  for (int i = 0; i < 31; i++) {
    if (l & (1 << i)) last = i;
  }
  for (int i = n - 1; i >= 0; i--) {
    long long num = l / (1 << i);
    tcost += num * cost[i];
    l -= num * (1 << i);
    if (l <= 0) {
      ans = min(ans, tcost);
    } else {
      ans = min(ans, cost[i] + tcost);
    }
  }
  cout << ans << "\n";
}
