Description

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
 

題目大意

現在有m種飲品，第i種飲品的價格為p[i]，容量為2i−1，現在輸出準備至少n容量的飲品至少需要多少錢。

解題思路

由於每相鄰兩種飲品的容量為2倍關係，所以一種情況下如果兩份 i 飲品的價格小於一份 i+1 飲品的價格，然後更新 i+1 種飲品的價格；另一種情況下如果 i+1 種飲品的價格小於 i 種飲品的價格，那麼變更新 i 種飲品的價格。以此遍歷兩次將每種飲品對應的價格更新至最小值。然後對總容量n的二進位制表示下的每一位進行判斷，如果當前位為1 ，那麼便購買當前位對應容量下的飲料，計算最終總價錢。

程式碼實現

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 37
ll p[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int m,n;
    cin>>m>>n;
    for(int i=0;i<m;i++)
        cin>>p[i];
    for(int i=1;i<m;i++)
        p[i]=min(p[i],p[i-1]*2);
    for(int i=m-2;i>=0;i--)
        p[i]=min(p[i],p[i+1]);
    int mark=0;
    for(int i=0;i<31;i++)
        if((1<<i)>n)
        {
            mark=i;
            break;
        }
    for(int i=m;i<=mark;i++)
        p[i]=p[i-1]*2;
    ll ans=0;
    for(int i=0;i<=mark;i++)
    {
        if(ans>p[i]) ans=p[i];
        if(n&(1<<i)) ans+=p[i];
    }
    cout<<ans<<endl;
    return 0;
}