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;
using ll = long long;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n, L;
  cin >> n >> L;
  vector<int> c(n);
  for (int i = 0; i < n; i++) cin >> c[i];
  vector<int> inds(n);
  iota(inds.begin(), inds.end(), 0);
  sort(inds.begin(), inds.end(), [&](int a, int b) {
    return ((long double)c[a] / (1 << a)) < ((long double)c[b] / (1 << b));
  });
  ll ans = 0, res = numeric_limits<ll>::max();
  for (int i = 0; i < n && L > 0; i++) {
    int cur = 1 << inds[i];
    if (cur <= L) {
      int times = L / cur;
      ans += 1ll * times * c[inds[i]];
      L -= times * cur;
    }
    res = min(res, ans + c[inds[i]]);
  }
  cout << min(res, ans) << '\n';
  return 0;
}
