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;
void preprocess(void) { return; }
long long f[32], t[32];
long long ans(long long x, long long y) {
  if (y <= 0) return 0;
  if (y >= t[x + 1]) return (long long)1000000007 * 1000000007;
  if (x == 0)
    return f[0];
  else
    return min(f[x] + ans(x - 1, y - t[x]), ans(x - 1, y));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  preprocess();
  long long n, l, x;
  t[0] = 1;
  cin >> n >> l;
  for (auto i = (0); i < n; i++) cin >> f[i];
  for (auto i = (1); i < 32; i++) {
    t[i] = t[i - 1] * 2;
  }
  for (int i = n - 2; i >= 0; i--) {
    f[i] = min(f[i], f[i + 1]);
  }
  for (auto i = (1); i < n; i++) {
    f[i] = min(f[i], f[i - 1] * 2);
  }
  for (auto i = (n); i < 32; i++) {
    f[i] = f[i - 1] * 2;
  }
  cout << ans(30, l);
}
