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 int mod = 1e9 + 7;
template <typename T>
void gi(T &ret) {
  ret = 0;
  char inp = getchar();
  int kl = 1;
  while (inp < '0' || inp > '9') {
    if (inp == '-') kl = -1;
    inp = getchar();
  }
  while ('0' <= inp && inp <= '9')
    ret = (ret << 3) + (ret << 1) + (inp - '0'), inp = getchar();
  if (kl < 1) ret = -ret;
}
const int bts = 30;
const long long inf = 1LL << 61LL;
int n, s;
long long val[bts + 5], sum = 0LL, ans = inf;
int main() {
  gi(n);
  gi(s);
  for (int i = 0; i < bts; i++) val[i] = inf;
  for (int i = 0; i < n; i++) gi(val[i]);
  for (int i = bts - 2; i >= 0; i--) val[i] = min(val[i], val[i + 1]);
  for (int i = 1; i < bts; i++) val[i] = min(val[i], val[i - 1] * 2LL);
  for (int i = bts - 1; i >= 0; i--) {
    if ((s >> i) & 1) sum += val[i];
    ans = min(ans, sum + val[i]);
  }
  ans = min(ans, sum);
  printf("%lld\n", ans);
  return 0;
}
