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 MAXN = 30 + 3;
struct Node {
  long long c, t;
  bool operator<(const Node &x) const {
    return (double)c / t < (double)x.c / x.t;
  }
} P[MAXN];
long long n, L, lft, ans, EXans = LLONG_MAX;
int main() {
  scanf("%I64d%I64d", &n, &L), lft = L;
  for (int i = 1; i <= n; i++) scanf("%I64d", &P[i].c), P[i].t = 1 << i - 1;
  sort(P + 1, P + 1 + n);
  for (int i = 1; i <= n; i++) {
    long long t0 = (long long)floor((double)lft / P[i].t),
              t1 = (long long)ceil((double)lft / P[i].t) * P[i].c,
              t2 = t0 * P[i].c;
    lft -= t0 * P[i].t, EXans = min(EXans, ans + t1), ans += t2;
  }
  printf("%I64d", EXans);
  return 0;
}
