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;
long long int c[35];
long long int m[35];
long long int solve(long long int L, int i) {
  long long int ans = (L / m[i]) * c[i];
  L = L % m[i];
  if (L != 0) return ans + min(c[i], solve(L, i - 1));
  return ans;
}
int main() {
  long long int n, L;
  cin >> n >> L;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    m[i] = pow(2, i);
  }
  for (int i = 0; i < n; i++) {
    long long int x = INT_MAX;
    for (int j = 0; j < n; j++) {
      if (j <= i) {
        x = min(x, (long long int)pow(2, i - j) * c[j]);
      } else {
        x = min(x, c[j]);
      }
    }
    c[i] = x;
  }
  int mini = -1;
  for (int i = 0; i < n - 1; i++) {
    if (c[i] < c[i + 1]) {
      mini = i;
      break;
    }
  }
  if (mini == -1) mini = n - 1;
  long long int ans = 0;
  int i = n - 1;
  ans = solve(L, i);
  cout << ans << endl;
  return 0;
}
