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.

題意：

有n家店鋪，每個店鋪有一種商品（無限個），每種商品有體積（2的i-1次方），每種商品有價格c[i],問至少買l體積的物品最少需要花費多少錢？

我們利用貪心的方法來解決。  我們優先選擇單位體積花費少的商品。

那麼我們的排序函式為

bool cmp(Node t1,Node t2)
{
    return t1.cost*t2.vol<t2.cost*t1.vol;
}

這是一個很常用的排序方法。

只是我們開始dfs，對於當前狀態，我們用最實惠的貨品一直買，買到買不下來了。還剩下零星的空間。

這個空間我們有兩個選擇，第一個就是再買這個商品，因為題目沒說不能超過給定的空間，只需要保證達到空間

耗費最少即可。所以我們可以這樣。第二個可能就是我們買第二實惠的商品，重複以上過程。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include<queue>
#include<map>
#include<math.h>
using namespace std;
struct Node
{
    long long vol;
    long long  cost;
}s[1100];
bool cmp(Node t1,Node t2)
{
    return t1.cost*t2.vol<t2.cost*t1.vol;
}
long long n,l;
long long dfs(long long left,long long cost,long long num)
{
    if(num>n)
        return 0;
    long long cout=left/s[num].vol; //判斷當前家需要購買幾瓶
    cost+=cout*s[num].cost;//加上花費
    left-=cout*s[num].vol;//計算剩餘體積
    if(left==0) return cost; //如果不剩餘了，則方案最優，直接返回
    return min(cost+s[num].cost,dfs(left,cost,num+1));//處理剩餘部分，再買一杯超出需要的量或者去下一家店，兩者去最少花費的那種。
}
int main(){
    s[1].vol=1;
    cin>>n>>l;
    cin>>s[1].cost;
    for(int i=2;i<=n;i++){
        cin>>s[i].cost;
        s[i].vol=2*s[i-1].vol;
    }
    sort(s+1,s+1+n,cmp);


//  ll tempm=0;
//  ll left=1;
//  ll temp;
    cout<<dfs(l,0,1)<<endl;
    return 0;
}