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;
vector<int> adia[100010];
bool ok = 1;
bool dfs(int nod) {
  int fii(0);
  if (adia[nod].size() == 0) return true;
  for (auto i : adia[nod]) fii += dfs(i);
  if (fii < 3) ok = 0;
  return false;
}
long long cost[40];
int main() {
  int n, l;
  cin >> n >> l;
  for (int i(0); i < n; i++) cin >> cost[i];
  for (int i(0); i <= 30; i++) {
    if (i + 1 >= n || 2 * cost[i] < cost[i + 1]) cost[i + 1] = 2 * cost[i];
  }
  for (int i(30); i > 0; i--)
    if (cost[i - 1] > cost[i]) cost[i - 1] = cost[i];
  long long ans(0);
  long long pos = (1ll << 60);
  for (int i(30); i >= 0; i--) {
    if (l & (1 << i)) ans += cost[i];
    pos = min(pos, ans + cost[i]);
  }
  cout << min(pos, ans);
  return 0;
}
