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;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <typename T = int>
inline T nxt() {
  char c = nc();
  T x = 0;
  int f = 1;
  for (; c > '9' || c < '0'; c = nc())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = nc())
    ;
  x *= f;
  return x;
}
const int N = 31;
long long cost[N];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int n = nxt(), l = nxt();
  for (int i = 0; i < (n); ++i) cost[i] = nxt();
  for (int i = 1; i < (n); ++i) cost[i] = min(2 * cost[i - 1], cost[i]);
  for (int i = n; i < (N); ++i) cost[i] = 2 * cost[i - 1];
  long long ans = 0;
  for (int i = 0; i < (N); ++i) {
    if (1LL * l & (1LL << i))
      ans += cost[i];
    else
      ans = min(cost[i], ans);
  }
  cout << ans << '\n';
}
