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;
const long long MX = (long long)9e18;
const int BITS = 32;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, l;
  cin >> n >> l;
  vector<long long> c(n);
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  long long ans = 0;
  long long mx = MX;
  vector<long long> best(BITS);
  best[0] = c[0];
  for (int i = 1; i < n; i++) {
    best[i] = min(c[i], 2 * best[i - 1]);
  }
  for (int i = n; i < BITS; i++) {
    best[i] = 2 * best[i - 1];
  }
  vector<int> bits;
  int y = l;
  while (y) {
    bits.push_back(y % 2);
    y /= 2;
  }
  for (int i = (int)bits.size() - 1; i >= 0; i--) {
    if (bits[i] == 0) {
      continue;
    }
    if (i + 1 > n - 1) {
      ans += best[i];
      continue;
    }
    long long mn = MX;
    for (int j = i + 1; j < n; j++) {
      mn = min(mn, c[j]);
    }
    mx = min(mx, ans + mn);
    ans += best[i];
  }
  ans = min(ans, mx);
  cout << ans << '\n';
  return 0;
}
