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;
unsigned long long cost_table[30][30];
unsigned long long n, need;
unsigned long long cost[30];
unsigned long long solve(int item, unsigned long long prev_waste,
                         unsigned long long need) {
  if (item < 0 || need <= 0) return 1000000000000000000;
  unsigned long long min_waste = 100000000000000000, curr_waste, avail, it;
  for (int i = 0; i < n; i++) {
    if (min_waste >= cost_table[i][item]) {
      it = i;
      min_waste = cost_table[i][item];
    }
  }
  unsigned long long prev_need = need;
  if (it < item) {
    it = item;
  }
  if (need >= (1 << it)) {
    curr_waste = (need / (1 << it)) * min_waste;
    need -= (need / (1 << it)) * (1 << it);
  } else {
    curr_waste = min_waste;
    need = 0;
  }
  if (need > 0) {
    return min(prev_waste + curr_waste + min_waste,
               solve(--item, prev_waste + curr_waste, need));
  } else
    return min(prev_waste + curr_waste, solve(--item, prev_waste, prev_need));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> need;
  for (int i = 0; i < n; i++) {
    cin >> cost[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j > i) {
        cost_table[i][j] = (cost_table[i][j - 1] * 2LL);
      } else {
        cost_table[i][j] = cost[i];
      }
    }
  }
  cout << solve(n - 1, 0, need) << endl;
  return 0;
}
