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 gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int n;
  long long int l;
  cin >> n >> l;
  long long int c[n + 1];
  for (int i = 1; i <= n; i++) cin >> c[i];
  for (int i = 1; i <= n - 1; i++)
    if (2 * c[i] <= c[i + 1]) c[i + 1] = 2 * c[i];
  long long int j = 1;
  int k = 1;
  int ar[33 + 1] = {0};
  long long int ans1 = 0, ans2;
  long long int ans, nm;
  while (j <= l) {
    if (j & l)
      ar[k++] = 1;
    else
      ar[k++] = 0;
    j = j << 1;
  }
  long long int p = 1, m = 0, sum = 0;
  for (int i = 1; i < n; i++)
    if (ar[i]) ans1 += c[i];
  ans2 = (l / pow(2, n - 1));
  ans1 = ans1 + c[n] * ans2;
  for (int i = 1; i <= n; i++) {
    if (ar[i]) {
      sum += c[i];
      m += p;
    } else {
      nm = l - m + p;
      if (i < n)
        ans = c[i];
      else
        ans = 0;
      for (int j = i + 1; j < n; j++)
        if (ar[j]) ans += c[j];
      ans2 = (long long int)(nm / pow(2, n - 1));
      ans = ans + c[n] * ans2;
      ans1 = min(ans1, ans);
    }
    p = p * 2;
  }
  ans2 = ceil((double)l / pow(2, n - 1));
  ans2 = c[n] * ans2;
  cout << min(ans1, ans2) << endl;
}
