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;
  }
  gcd(b % a, a);
}
int main() {
  long long int t, m, o, e, a, k, b, l, r, c;
  long long int n;
  cin >> n >> l;
  std::vector<long long int> v2;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v2.push_back(x);
  }
  for (long long int i = 0; i < n - 1; i++) {
    v2[i + 1] = min(v2[i + 1], 2 * v2[i]);
  }
  long long int ans = 9223372036854775801;
  long long int sum = 0;
  for (long long int i = n - 1; i >= 0; i--) {
    long long int need = l / (1 << i);
    sum += need * v2[i];
    l -= need << i;
    ans = min(ans, sum + (l > 0) * v2[i]);
  }
  cout << ans;
  return 0;
}
