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;
struct www {
  int c, l;
  double s;
} f[1000];
int n, l, i, p;
long long res, nw;
bool cmp(const www &a, const www &b) { return a.s < b.s; }
int main() {
  cin >> n >> l;
  for (i = 1; i <= n; i++) cin >> f[i].c;
  f[1].l = 1;
  for (i = 2; i <= n; i++) f[i].l = f[i - 1].l * 2;
  for (i = 1; i <= n; i++) f[i].s = 1.0 * f[i].c / f[i].l;
  sort(f + 1, f + n + 1, cmp);
  res = 1e18;
  nw = 0;
  for (i = 1; i <= n; i++) {
    p = l / f[i].l;
    nw = nw + 1ll * p * f[i].c;
    if (l % f[i].l == 0) {
      if (nw < res) res = nw;
      break;
    }
    if (nw + f[i].c < res) res = nw + f[i].c;
    l = l % f[i].l;
  }
  cout << res << endl;
  return 0;
}
