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;
int main(void) {
  cin.sync_with_stdio(0);
  cin.tie(NULL);
  cout << fixed << setprecision(25);
  long long n, L;
  cin >> n >> L;
  vector<long long> c(n);
  for (int i = 0; i < n; i++) cin >> c[i];
  vector<long long> c_lit(n);
  for (int i = 0; i < n; i++) {
    c_lit[i] = c[i] << (n - 1 - i);
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < i; j++) {
      if (c_lit[i] >= c_lit[j]) {
        c[i] = -1;
        break;
      }
    }
  }
  int i = n - 1;
  long long pow = 1;
  for (int j = 0; j < i; j++) {
    pow *= 2;
  }
  long long prev_ans = 5000000000000000000;
  long long cur_cost = 0;
  long long cur_lit = L;
  while (i >= 0) {
    if (c[i] == -1) {
      pow /= 2;
      i--;
      continue;
    }
    cur_cost = cur_cost + (cur_lit / pow) * c[i];
    cur_lit = cur_lit % pow;
    prev_ans = min(prev_ans, cur_cost + c[i]);
    i--;
    pow /= 2;
  }
  cout << min(cur_cost, prev_ans);
  return 0;
}
