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;
long long p[35];
int main() {
  int n, l;
  long long ret = -1;
  scanf("%d%d", &n, &l);
  for (int i = 0; i < n; i++) scanf("%lld", &p[i]);
  for (int i = 1; i < 31; i++)
    if (p[i] == 0 or p[i] > p[i - 1] * 2) p[i] = p[i - 1] * 2;
  for (int i = 30; (1 << i) >= l; i--)
    if (ret < 0 or ret > p[i]) ret = p[i];
  string bin = "";
  while (l) {
    bin += (l % 2 + '0');
    l /= 2;
  }
  long long t = 0;
  for (int i = bin.length() - 1; i >= 0; i--) {
    if (ret == -1 or ret > p[i] * (bin[i] - '0' + 1) + t)
      ret = p[i] * (bin[i] - '0' + 1) + t;
    t += (bin[i] - '0') * p[i];
  }
  ret = min(ret, t);
  printf("%lld\n", ret);
}
