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;
using ll = long long;
using ld = long double;
const ll inf = 1e18;
ll a[35], dp[35];
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed;
  ll n, L;
  cin >> n >> L;
  memset(dp, 0x3f, sizeof(dp));
  for (int i = 0; i < n; i++) cin >> a[i];
  dp[0] = a[0];
  for (int i = 1; i < 31; i++) {
    if (i < n)
      dp[i] = min(dp[i - 1] * 2, a[i]);
    else
      dp[i] = dp[i - 1] * 2;
  }
  for (int j = 30; j >= 0; j--) {
    dp[j] = min(dp[j], dp[j + 1]);
  }
  vector<int> V;
  ll tmp = L;
  while (tmp) {
    V.push_back(tmp % 2);
    tmp /= 2;
  }
  ll res = dp[((int)V.size())], cnt = 0;
  for (int i = ((int)V.size()) - 1; i >= 0; i--) {
    if (!V[i]) {
      res = min(res, cnt + dp[i]);
    } else {
      cnt += dp[i];
    }
  }
  cout << min(res, cnt) << endl;
  return 0;
}
