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;
const int MAXN = (int)(2e4) + 2;
const int MOD = (int)(1e9) + 7;
int n, L, a[33];
pair<double, pair<int, int> > p[33];
long long ans = 2e18;
long long rec(int X, int pos = 1) {
  if (pos > n) return 2e18;
  long long res = 2e18;
  int lg = (1 << (p[pos].second.first - 1));
  res =
      min((X + lg - 1) / lg * 1ll * p[pos].second.second,
          X / lg * 1ll * p[pos].second.second + rec(X - X / lg * lg, pos + 1));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> L;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p[i] = make_pair(a[i] * 1.0 / (1 << (i - 1)), make_pair(i, a[i]));
  }
  sort(p + 1, p + 1 + n);
  cout << rec(L);
  return 0;
}
