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 expo(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b /= 2;
  }
  return res;
}
int main() {
  int n, i, len, j;
  long long L, temp;
  cin >> n >> L;
  temp = L;
  vector<long long> c(n);
  for (i = 0; i < n; i++) cin >> c[i];
  string s = "";
  while (temp) {
    s += ('0' + temp % 2);
    temp /= 2;
  }
  for (i = 1; i < n; i++) {
    c[i] = min(c[i], 2 * c[i - 1]);
  }
  reverse(s.begin(), s.end());
  len = s.size();
  long long ans = 1e18, need, sum = 0;
  temp = L;
  for (i = n - 1; i >= 0; i--) {
    need = (temp / expo(2, i));
    sum += need * c[i];
    temp -= (need * expo(2, i));
    ans = min(ans, sum + (temp > 0) * c[i]);
  }
  ans = min(ans, sum);
  for (i = 0; i < len; i++) {
    if (s[i] == '1') {
      ans = min(ans, sum);
    } else {
      temp = L;
      temp += expo(2, i);
      long long sum2 = 0;
      for (j = n - 1; j >= 0; j--) {
        need = (temp / expo(2, j));
        sum2 += need * c[j];
        temp -= (need * expo(2, j));
      }
      ans = min(ans, sum2);
    }
  }
  cout << ans << '\n';
}
