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;
struct Bottle {
  int64_t c, p;
  double r;
};
bool cmp(const Bottle &lhs, const Bottle &rhs) {
  if (lhs.r == rhs.r) return lhs.c > rhs.c;
  return lhs.r < rhs.r;
}
vector<Bottle> b;
int64_t n;
int64_t f(int64_t i, int64_t m) {
  int64_t t = m / b[i].c;
  m -= t * b[i].c;
  int64_t p = t * b[i].p;
  if (m == 0) return p;
  if (i == n - 1) return p + b[i].p;
  int64_t tmp = f(i + 1, m);
  p += (tmp < b[i].p ? tmp : b[i].p);
  return p;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int64_t m;
  cin >> n >> m;
  for (int64_t i = 0; i < n; i++) {
    int64_t x;
    cin >> x;
    b.push_back({(1 << i), x, double(x) / (1 << i)});
  }
  sort(b.begin(), b.end(), cmp);
  cout << f(0, m) << endl;
  return 0;
}
