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;
bool rev(long long x, long long y) { return x > y; }
void input_arr(long long a[], long long n) {
  for (long long i = 0; i < n; i += 1) cin >> a[i];
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return a.second < b.second;
}
bool sortinrev(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
void print_vec(vector<long long> &v) {
  for (auto i : v) cout << i << " ";
  cout << '\n';
}
bool sortinrevbysec(const pair<long long, long long> &a,
                    const pair<long long, long long> &b) {
  return (a.second > b.second);
}
void print_arr(long long a[], long long n) {
  for (long long i = 0; i < n; i += 1) cout << a[i] << " ";
  cout << '\n';
}
template <typename... T>
void input(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void print(T &&...args) {
  ((cout << args << '\n'), ...);
}
template <typename... T>
void deb(T &&...args) {
  cout << "~~ ";
  ((cout << args << " "), ...);
  cout << '\n';
}
void stand8th() {
  long long n, l;
  input(n, l);
  long long a[n];
  input_arr(a, n);
  for (long long i = 1; i < n; i += 1) {
    a[i] = min(a[i], a[i - 1] * 2);
  }
  long long ans = INT64_MAX;
  long long summ = 0;
  for (long long i = n - 1; i >= 0; i--) {
    if (pow(2, i) >= l) {
      ans = min(ans, summ + a[i]);
    } else {
      long long cnt = l / (long long)(pow(2, i));
      summ += cnt * a[i];
      l -= cnt * (long long)(pow(2, i));
      if (l == 0) {
        ans = min(ans, summ);
      } else {
        ans = min(ans, summ + a[i]);
      }
    }
  }
  print(ans);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long tin = 1;
  while (tin--) {
    stand8th();
  }
  return 0;
}
