codeforces Hello 2018 C. Party Lemonade（DP+思维）

C. Party Lemonade

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

思路：

dp[0]表示20 体积的最优价格，以此类推，

dp[i]表示2i 体积的最优价格。

然后二进制拆分需要买的体积，则对应的二进制位为1则取dp[i]的最优价格，为0则取当前价钱>总商品的价格，则取总商品的价格。

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
const int N=35;
int main()
{
ll dp
;
ll n,L;
scanf("%lld%lld",&n,&L);
for(int i=0;i<n;i++)
{
scanf("%lld",&dp[i]);
}
for(int i=1;i<n;i++)
{
dp[i]=min(dp[i],dp[i-1]*2);
}
for(int i=n-2;i>=0;i--)
{
dp[i]=min(dp[i+1],dp[i]);
}
for(int i=n;i<31;i++)
{
dp[i]=dp[i-1]<<1;
//printf("%lld %lld \n",dp[i],dp[i-1]);
}
ll sum=0;
for(int i=0;i<31;i++)
{
if(L&(1LL<<i))sum+=dp[i];
else if(sum>dp[i])sum=dp[i];
}
printf("%lld\n",sum);
return 0;
}