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 n, L, k;
long long ans, c[34];
inline long long dfs(int x) {
  int pos = 34;
  for (int i = 0; i <= 30; i++)
    if (x < (1 << i)) {
      pos = i - 1;
      break;
    }
  if (x == (1 << pos)) return c[pos];
  long long rest = x - (1 << pos);
  long long sum = dfs(rest);
  int pos1 = 34;
  for (int i = 0; i <= 30; i++)
    if (rest < (1 << i)) {
      pos1 = i - 1;
      break;
    }
  for (int i = pos1 + 1; i <= pos; i++) sum = min(sum, c[i]);
  return c[pos] + sum;
}
int main() {
  scanf("%d%d", &n, &L);
  for (int i = 0; i < n; i++) scanf("%lld", &c[i]);
  for (int i = 1; i < n; i++) c[i] = min(c[i], c[i - 1] + c[i - 1]);
  for (int i = n; i <= 30; i++) c[i] = c[i - 1] + c[i - 1];
  k = 34;
  for (int i = 0; i <= 30; i++)
    if (L < (1 << i)) {
      k = i;
      break;
    }
  ans = dfs(L);
  for (int i = k; i <= 30; i++) ans = min(ans, c[i]);
  printf("%I64d\n", ans);
  return 0;
}
