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 bbuy(long long* c, long long l, int n);
int main() {
  long long n, l;
  cin >> n >> l;
  long long c[n];
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
  }
  long long bp = c[0];
  for (int i = 0; i < n; ++i) {
    bp *= 2;
    bp = min(bp, c[i]);
    if (bp != c[i]) c[i] = -1;
  }
  long long bcost = 0;
  bcost += bbuy(c, l, n);
  cout << bcost;
  return 0;
}
long long bbuy(long long* c, long long l, int n) {
  int i = int(log2(l));
  long long gcost = 0;
  long long cost = 0, buy;
  long long prev = l;
  while (i >= n || c[i] == -1) i--;
  buy = l / pow(2, i);
  l -= buy * pow(2, i);
  cost = buy * c[i];
  if (l > 0) cost += bbuy(c, l, n);
  for (int j = i + 1; j < n; ++j) {
    l = prev;
    if (c[j] == -1) continue;
    buy = (l / pow(2, j)) + 1;
    if (cost > buy * c[j]) {
      cost = buy * c[j];
      l -= buy * pow(2, j);
    }
  }
  gcost += cost;
  return gcost;
}
