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>
const int Max = 100000;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  int n;
  long long l;
  long long c[40];
  long long value[40];
  value[0] = 1;
  while (cin >> n >> l) {
    long long lubu = 0;
    long long lubua = 10000000001;
    cin >> c[0];
    for (int i = 1; i < n; i++) {
      int temp;
      cin >> temp;
      long long min = temp;
      int ans = 2;
      for (int j = i - 1; j >= 0; j--) {
        if (min > c[j] * ans) {
          min = c[j] * ans;
        }
        ans = ans * 2;
      }
      c[i] = min;
      value[i] = value[i - 1] * 2;
    }
    for (int i = 0; i < n; i++) {
      if (value[i] >= l) {
        if (c[i] < lubua) {
          lubua = c[i];
        }
      }
    }
    long long l1 = l;
    long long lubub;
    lubub = 66666666666666666;
    int ww = n - 1;
    while (l >= 0) {
      if (l == 0) break;
      if (value[ww] > l) {
        for (int i = 0; i < n; i++) {
          if (value[i] <= l)
            ww = i;
          else if (value[i] > l) {
            break;
          }
        }
      }
      int temp = l / value[ww];
      lubu = lubu + c[ww] * temp;
      l = l % value[ww];
      for (int i = n - 1; i >= 0 && value[i] > l; i--) {
        if (value[i] > l) {
          if (lubub > lubu + c[i]) {
            lubub = lubu + c[i];
          }
        }
      }
    }
    unsigned long long answer;
    if (lubua == 10000000001) {
      if (lubub != 66666666666666666) {
        answer = min(lubub, lubu);
      } else
        answer = lubu;
    } else {
      if (lubub != 66666666666666666) {
        answer = min(lubub, min(lubu, lubua));
      } else
        answer = min(lubu, lubua);
    }
    cout << answer << endl;
  }
  return 0;
}
