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 power(long long int a, long long int b, long long int n) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % n;
    a = (a * a) % n;
    b /= 2;
  }
  return ans;
}
pair<long long int, long long int> c[32];
map<pair<long long int, long long int>, long long int> m;
long long int get(long long int l, long long int n) {
  if (l <= 0) return 0;
  double m = 10000000000000000;
  int p;
  for (int i = 0; i <= n; i++) {
    double x = ((double)c[i].first) / c[i].second;
    if (x < m) {
      m = x;
      p = i;
    }
  }
  if (p == 0) return l * c[0].first;
  long long int a = (l / c[p].second) * c[p].first;
  return min(a + c[p].first,
             a + get(l - (l / c[p].second) * c[p].second, n - 1));
}
signed main() {
  long long int n, l;
  scanf("%lld", &(n));
  scanf("%lld", &(l));
  for (int i = 0; i < n; i++) {
    scanf("%lld", &(c[i].first));
    c[i].second = pow(2, i);
  }
  cout << get(l, n - 1);
  return 0;
}
