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>
int n;
long long c[33];
std::map<std::pair<int, long long>, long long> memo;
long long dfs(int i, long long L) {
  long long q = 1LL << i;
  if (L <= 0) return 0;
  std::pair<int, long long> key(i, L);
  if (memo.count(key)) return memo[key];
  long long Cm = 4000000000000000000LL;
  for (int k = 0; k < n; k++) {
    long long C = c[k] << std::max((i - k), 0);
    if (C < Cm) Cm = C;
  }
  long long x = L / q, S = Cm * x;
  L %= q;
  if (i && L) S += std::min(Cm, dfs(i - 1, L));
  return memo[key] = S;
}
int main(int argc, char** argv) {
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(0);
  std::cout << std::fixed << std::setprecision(6);
  int L;
  std::cin >> n >> L;
  for (int i = 0; i < n; i++) std::cin >> c[i];
  std::cout << dfs(n - 1, L) << std::endl;
  return 0;
}
