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;
int n, l;
unsigned long long dp[65];
int arr[65];
unsigned long long best(int i) {
  if (i == 64) return 1e18;
  unsigned long long& ret = dp[i];
  if (ret != -1) return ret;
  unsigned long long sol1 = 1e18, sol2 = 1e18;
  if (i < n) sol1 = arr[i];
  if (i > 0) sol2 = best(i - 1) * 2;
  return ret = min(sol1, sol2);
}
unsigned long long dpNorm(int i) {
  if (i == 64) return 1e19;
  dp[i] = min(best(i), dpNorm(i + 1));
  return dp[i];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n >> l;
  for (int i = 0; i < n; i++) cin >> arr[i];
  memset(dp, -1, sizeof(dp));
  best(0);
  dpNorm(0);
  unsigned long long ans = 0;
  unsigned long long i;
  for (i = 0; i < n || (1 << i) <= l; i++) {
    if ((l & (1 << i)) != 0) {
      ans += best(i);
    } else {
      ans = min(ans, best(i));
    }
  }
  cout << min(ans, best(i));
}
