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;
unsigned long long n, l, a[45], ans;
vector<unsigned long long> v;
unsigned long long fp(int x, int y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  unsigned long long ret = fp(x, y / 2);
  ret *= ret;
  if (y % 2 == 1) ret *= x;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> l;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    if (a[i] * 2 < a[i + 1]) a[i + 1] = a[i] * 2;
  }
  for (int i = n; i <= 30; i++) a[i] = 2 * a[i - 1];
  unsigned long long cur = l;
  for (int i = 29; i >= 0; i--) {
    if (fp(2, i) > cur) {
      v.push_back(ans + a[i]);
      continue;
    }
    cur -= fp(2, i);
    ans += a[i];
  }
  for (int i = 0; i <= 30; i++) {
    if (fp(2, i) >= l) ans = min(ans, a[i]);
  }
  for (int i = 0; i < v.size(); i++) ans = min(ans, v[i]);
  cout << ans << endl;
  return 0;
}
