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 T, m, n;
long long ans;
long long L;
long long c[35];
bool f[35];
int main() {
  memset(f, false, sizeof(f));
  scanf("%d%lld", &n, &L);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
  }
  bool flag = true;
  while (flag) {
    flag = false;
    for (int i = n; i > 1; i--) {
      if (c[i] < c[i - 1]) {
        flag = true;
        c[i - 1] = c[i];
        f[i - 1] = true;
      }
    }
    for (int i = 1; i < n; i++) {
      if (c[i] * 2 < c[i + 1]) {
        flag = true;
        c[i + 1] = 2 * c[i];
      }
    }
  }
  ans = 0;
  int cnt = 0;
  while (L > 0) {
    cnt++;
    if (ans > c[cnt]) ans = c[cnt];
    if (cnt >= n) {
      ans += c[cnt] * L;
      L = 0;
      cnt++;
    } else {
      if (L % 2 > 0) {
        ans += c[cnt];
      }
      L /= 2;
    }
  }
  cnt++;
  while (cnt <= n) {
    if (ans > c[cnt]) ans = c[cnt];
    cnt++;
  }
  printf("%lld\n", ans);
  return 0;
}
