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>
#pragma GCC optimize("Ofast", "unroll-loops")
using namespace std;
const long long inf = numeric_limits<long long>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
long long n, l;
vector<long long> c;
bool checkbit(long long n, long long p) { return n & (1 << p); }
long long setbit(long long n, long long p) { return n | (1 << p); }
inline void solve() {
  for (long long i = 1; i < n; i++) {
    (c[i]) = min((c[i]), (2 * c[i - 1]));
  }
  long long ans = inf;
  long long sum = 0;
  for (long long i = n - 1; i >= 0; i--) {
    long long m = l / (1ll << i);
    sum += m * c[i];
    l -= m * (1ll << i);
    long long d = 0;
    if (l > 0) {
      d = c[i];
    }
    (ans) = min((ans), (sum + d));
  }
  cout << ans << "\n";
}
inline void read() {
  cin >> n >> l;
  c.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> c[i];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
