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;
const int N = 33;
const long long LINF = (long long)1e18;
int n;
long long L, c[N];
long long v[N];
long long dfs(int x, long long R) {
  if (R <= 0) return 0;
  if (x < 0 && R > 0) return LINF;
  long long res = dfs(x - 1, R - (1 << x)) + c[x];
  if ((1 << x) > R) res = min(res, dfs(x - 1, R));
  return res;
}
int main() {
  scanf("%d%I64d", &n, &L);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &c[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      c[j] = min(c[j], c[i] << (j - i));
    }
  }
  long long num = L >> (n - 1);
  long long base = num * c[n - 1];
  L -= num << (n - 1);
  long long ans = dfs(n - 1, L);
  ans += base;
  printf("%I64d\n", ans);
  return 0;
}
