CoderForces-913-C

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 2^i - 1 liters and costs c_i 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, ..., c_n (1 ≤ c_i ≤ 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.

Example

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.

题解：

因为 2n−1×2=2n2n−1×2=2n ，所以我们可以想到小物品组成大物品是否可以带来更小的花费。

于是从小到大扫一遍计算出组成当前大小为 2i2i 所需要的最小花费，记为 aiai 。

然后针对大小 LL ，我们可以将其转换为二进制，从高位往低位开始枚举，

用 nownow 记录已访问的高位中所需要的花费，若当前位为 11 ， now+=a[i]now+=a[i] ，因为我们不能通过这一位组合出大小大于 LL 的货物，

若当前位为 00 ，记录 now+a[i]now+a[i] ，因为此时我们只需要将该位填充为 11 即可组出大于 LL 的货物。

然后找最小值即可。

AC代码为：

#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
using ll = long long;
ll cost[50];
int main() 
{
int n, l;
cin >> n >> l;
for (int i = 1; i <= n; ++i)
{
cin >> cost[i];
}


for (int i = 1; i <= 30; ++i) 
{
if (i + 1 > n || cost[i] * 2 < cost[i + 1])
cost[i + 1] = cost[i] * 2;
}


for (int i = 30; i >= 1; --i) 
{
if (cost[i + 1] < cost[i])
cost[i] = cost[i + 1];
}
ll ans = 0;
ll res = 1e18;
for (int i = 30; i >= 0; --i)
{
if (l & (1ll << i))
ans += cost[i + 1];


res = min(ans + cost[i + 1], res);
}
cout << min(res, ans) << endl;
}