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 maxn = 9223000000000000000;
struct DATA {
  long long v;
  long long c;
  long double p;
};
long long n, l;
DATA data[31];
bool cmp(DATA x, DATA y) { return x.p < y.p; }
void init() {
  scanf("%lld%lld", &n, &l);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &data[i].c);
    data[i].v = pow(2, i - 1);
    data[i].p = (long double)data[i].c / data[i].v;
  }
  sort(data + 1, data + n + 1, cmp);
}
long long process(long long left) {
  long long ans = maxn;
  if (left == 0) return 0;
  for (int i = 1; i <= n; i++) {
    if (data[i].v >= left) ans = min(ans, data[i].c);
  }
  int id = -1;
  for (int i = 1; i <= n; i++)
    if (data[i].v < left) {
      id = i;
      break;
    }
  if (id != -1) {
    long long sum = 0;
    long long k = left / data[id].v;
    sum = k * data[id].c + process(left - k * data[id].v);
    ans = min(ans, sum);
  }
  return ans;
}
int main() {
  init();
  printf("%lld", process(l));
  return 0;
}
