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;
const long long M = 1e9 + 7;
const long long INF = 1e9;
inline long long pwr(long long base, long long n, long long m) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans = (ans * base) % m;
    base = (base * base) % m;
    n /= 2;
  }
  return ans;
}
long long n, c[32];
long long rec(long long vol) {
  if (vol == 0) return 0;
  int ind = 0;
  long long cc;
  long long extra = (1LL << 62);
  for (int i = 0; i < n; i++) {
    if ((1 << i) <= vol) {
      ind = i;
      cc = (1 << i);
    } else
      extra = min(extra, c[i]);
  }
  return min(extra, c[ind] * (vol / cc) + rec(vol % cc));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long l;
  cin >> n >> l;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    if (i > 0) c[i] = min(c[i], c[i - 1] * 2);
  }
  cout << rec(l);
  return 0;
}
