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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int a[32];
long long int cal(long long int x, long long int l1) {
  long long int i;
  if (l1 == 0) return 0;
  bool f = 0;
  for (i = x - 1; i >= 0; i--) {
    long long int p = (1 << x) / (1 << i);
    if (a[i] * p < a[x]) {
      f = 1;
      break;
    }
  }
  if (f == 1) return cal(x - 1, l1);
  long long int po = (1 << x);
  long long int k = l1 / po;
  long long int baki = l1 % po;
  long long int r1 = cal(x - 1, baki);
  long long int r2 = k * a[x];
  long long int r3 = min(r1, a[x]);
  return r3 + r2;
}
int main() {
  long long int n, l, i;
  while (cin >> n >> l) {
    for (long long int i = 0; i < n; i++) cin >> a[i];
    long long int ans = cal(n - 1, l);
    cout << ans << endl;
  }
  return 0;
}
