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 cost[33];
long long int dp[35];
bool flag[35];
long long int n, L;
int main() {
  cin >> n >> L;
  for (int i = 0; i < n; i++) cin >> cost[i];
  for (int i = 0; i < 33; i++) {
    if (i < n) {
      dp[i] = cost[i];
      if (i > 0) dp[i] = min(dp[i], 2LL * dp[i - 1]);
    } else {
      dp[i] = 2LL * dp[i - 1];
    }
  }
  long long int minn = dp[32];
  for (int i = 32; i >= 0; i--) {
    if (minn < dp[i]) {
      dp[i] = minn;
      flag[i] = 1;
    }
    minn = min(minn, dp[i]);
  }
  long long int res2 = 1000000000000000000;
  long long int res = 0;
  long long int temp = 0;
  while (L > 0) {
    long long int x = log2(L);
    temp = res + dp[x + 1];
    res2 = min(res2, temp);
    res += dp[x];
    L -= (1LL << x);
  }
  res = min(res, res2);
  cout << res << endl;
  return 0;
}
