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* num_to_bit(long long n, int k) {
  stack<int> second;
  static int a[65] = {0};
  int i = 0;
  while (n > 0) {
    int m = n % k;
    a[i] = m;
    i++;
    second.push(m);
    n /= k;
  }
  return a;
}
int bit(long long n) {
  int c = 0;
  if (n == 0)
    return 1;
  else {
    while (n != 0) {
      c++;
      n /= 2;
    }
    return c;
  }
}
void ans_nikalo() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, l;
  cin >> n >> l;
  long long a[n];
  vector<pair<long long, long long>> v;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    if (2 * a[i] < a[i + 1]) {
      a[i + 1] = min(2 * a[i], a[i + 1]);
    }
  }
  long long ans = LLONG_MAX;
  long long sum = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long m = pow(2, i);
    long long q = l / m;
    long long r = l % m;
    l -= q * m;
    sum += a[i] * q;
    ans = min(ans, sum + (l > 0) * a[i]);
  }
  cout << ans;
  return;
}
int main() {
  ans_nikalo();
  return 0;
}
