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>
const unsigned long long inf = (1LL << 63);
using namespace std;
unsigned long long p[40], c[40];
double pc[40];
int main() {
  int n, L;
  cin >> n >> L;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    p[i] = 1 << i;
    pc[i] = (double)c[i] / p[i];
  }
  unsigned long long flag = L, sum = 0, ans = inf;
  while (flag > 0) {
    int i;
    for (i = n - 1; i >= 0 && p[i] > flag; i--) {
      ans = min(ans, sum + c[i]);
    }
    double s = inf;
    int id;
    for (int j = 0; j <= i; j++) {
      if (s >= pc[j]) {
        s = pc[j];
        id = j;
      }
    }
    sum += c[id] * (flag / p[id]);
    flag %= p[id];
    if (flag == 0) ans = min(ans, sum);
  }
  cout << ans << endl;
}
