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;
struct cww {
  cww() {
    if (1) {
      ios::sync_with_stdio(false);
      cin.tie(0);
    }
  }
} star;
template <typename T>
inline bool chmin(T &l, T r) {
  bool a = l > r;
  if (a) l = r;
  return a;
}
template <typename T>
inline bool chmax(T &l, T r) {
  bool a = l < r;
  if (a) l = r;
  return a;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &it : v) is >> it;
  return is;
}
class range {
 private:
  struct I {
    int x;
    int operator*() { return x; }
    bool operator!=(I &lhs) { return x < lhs.x; }
    void operator++() { ++x; }
  };
  I i, n;

 public:
  range(int n) : i({0}), n({n}) {}
  range(int i, int n) : i({i}), n({n}) {}
  I &begin() { return i; }
  I &end() { return n; }
};
int N;
long long L;
int main() {
  while (cin >> N >> L) {
    vector<long long> c(100, 1e18);
    for (int i : range(N)) {
      cin >> c[i];
    }
    for (int i : range(1, 100)) {
      chmin(c[i], 2 * c[i - 1]);
    }
    long long ret = 0;
    int id = 0;
    while (L) {
      chmin(ret, c[id]);
      if (L & 1) ret += c[id];
      L >>= 1;
      id++;
    }
    for (int i : range(id, 50)) {
      chmin(ret, c[i]);
    }
    cout << ret << endl;
  }
  return 0;
}
