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;
long long int mod = 1e9 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  if (a < b) {
    long long int t = a;
    a = b;
    b = t;
  }
  return (a * b) / gcd(a, b);
}
long long int modinv(long long int x) {
  int y = mod - 2;
  long long int res = 1;
  while (y > 0) {
    if (y % 2 == 1) res = res * x % mod;
    y /= 2;
    x = x * x % mod;
  }
  return res;
}
bool cmp(pair<int, long long int> &a, pair<int, long long int> &b) {
  double r1 = (double)a.first / (double)a.second;
  double r2 = (double)b.first / (double)b.second;
  return r1 < r2;
}
void solve() {
  int n;
  long long int l;
  cin >> n >> l;
  pair<int, long long int> b[n];
  for (int i = 0; i < n; i++) {
    cin >> b[i].first;
    b[i].second = pow(2, i) + 0.5;
  }
  sort(b, b + n, cmp);
  long long int D[n + 1][2];
  memset(D, 0, sizeof D);
  long long int mi = 1e18;
  for (int i = 1; i <= n; i++) {
    int c = 0;
    if (l % b[i - 1].second != 0) c = 1;
    D[i][0] = (l / b[i - 1].second + c) * b[i - 1].first + D[i - 1][1];
    D[i][1] = (l / b[i - 1].second) * b[i - 1].first + D[i - 1][1];
    l = l % b[i - 1].second;
    mi = min(mi, D[i][0]);
  }
  cout << mi;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
    cout << endl;
  }
}
