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;
void Genawy() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}
int n;
long long l;
vector<long long> cost;
vector<bool> used;
long long solve(long long cur) {
  double mn = 1e14;
  int u = -1;
  for (int i = 0; i < n; i++) {
    if (used[i]) continue;
    if (cost[i] * 1.0 / (1LL << i) < mn) {
      mn = cost[i] * 1.0 / (1LL << i);
      u = i;
    }
  }
  if (u == -1) return 1e14;
  used[u] = true;
  long long ret = cur / (1LL << u) * cost[u];
  long long rem = cur % (1LL << u);
  if (rem == 0) return ret;
  long long ret1 = cost[u];
  ret1 = min(ret1, solve(rem));
  return ret + ret1;
}
int main() {
  Genawy();
  cin >> n;
  cin >> l;
  cost.resize(n);
  used.resize(n);
  for (int i = 0; i < n; i++) cin >> cost[i];
  cout << solve(l) << endl;
  return 0;
}
