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 n, m;
long long a[100];
long long ans = 1e18;
long long po[100];
map<long long, long long> ma;
long long fun(long long left1) {
  long long ans1 = 1e18;
  if (left1 <= 0) {
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    long long left = left1;
    long long temp = 0;
    if (po[i] >= left)
      left = 0, temp = a[i];
    else if (left % po[i] == 0) {
      temp = left / po[i];
      temp *= a[i];
      left = 0;
    } else {
      temp = left / po[i];
      temp *= a[i];
      left = left % po[i];
    }
    if (ma[left] != 0)
      ans1 = min(ans1, temp + ma[left]);
    else {
      ma[left] = fun(left);
      ans1 = min(ans1, temp + ma[left]);
    }
  }
  return ans1;
}
int main() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> a[i];
  po[0] = 1;
  for (long long i = 1; i <= 30; i++) {
    po[i] = po[i - 1] * 2;
  }
  for (long long i = 0; i < n; i++) {
    long long left = m;
    long long temp;
    if (po[i] >= left)
      left = 0, temp = a[i];
    else if (left % po[i] == 0) {
      temp = left / po[i];
      temp *= a[i];
      left = 0;
    } else {
      temp = left / po[i];
      temp *= a[i];
      left = left % po[i];
    }
    ans = min(ans, temp + fun(left));
  }
  cout << ans;
  return 0;
}
