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;
long long calculate_cost(vector<long long> m, vector<long long> pow,
                         long long L);
int main() {
  long long N, L;
  cin >> N >> L;
  vector<long long> m;
  vector<long long> pow;
  for (long long i = 0; i < N; i++) {
    long long r;
    cin >> r;
    m.push_back(r);
  }
  pow.push_back(1);
  for (long long i = 0; i < N; i++) {
    pow.push_back(pow[pow.size() - 1] * 2);
  }
  cout << calculate_cost(m, pow, L);
  return 0;
}
long long calculate_cost(vector<long long> m, vector<long long> pow,
                         long long L) {
  long long ans = 0;
  double max_nom = pow[0] / m[0];
  long long max_ind = 0;
  for (long long i = 0; i < m.size(); i++) {
    if ((double)pow[i] / m[i] > max_nom) {
      max_nom = (double)pow[i] / m[i];
      max_ind = i;
    }
  }
  ans = (L / pow[max_ind]) * m[max_ind];
  L -= (L / pow[max_ind]) * pow[max_ind];
  if (L == 0) return ans;
  long long ans2 = ans + m[max_ind];
  m.erase(m.begin() + max_ind);
  pow.erase(pow.begin() + max_ind);
  long long ans1 = ans + calculate_cost(m, pow, L);
  return min(ans1, ans2);
}
