code force Party Lemonade

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.

题意：你要买水，至少l升，水有n种规格，第二行n个数字，表示第i种水需要ci元，第i种水的重量是2^(i-1)升，求你至少花费x元钱至少买l升水。

思路：假设只能买l升水，不能多买也不能少买。处理一下数据，将买1，2，4...升水最少需要多少钱求出来，（if(ci>2*c(i-1))那么,ci=2*ci-1) 那么需要买的水的重量可以用售卖的水一个二进制的字符串来表示。（此时x+=((l>>i)&1)*ci;)

由于此题可以多买，也就是说最后这个二进制的字符串的值会大于等于之前二进制的字符串的值，(例如13(1101),可以买14(1110))这个时候只要在前面的式子再加个判断条件即可（if(x>ci) x=ci;x+=((l>>i)&1)*ci;)

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int n,m;
    long long ans=0,c[45];
   // printf("%d",(int)pow(2,64));
    scanf("%d%d",&n,&m);
    for(int i=1;i<32;i++)
    {
        if(i==1)
            scanf("%lld",&c[i]);
        else if(i<=n)
        {
            scanf("%lld",&c[i]);
            c[i]=min(c[i],c[i-1]*2);
        }
        else c[i]=c[i-1]*2;
    }
    for(int i=1;i<32;i++)
    {
        ans=min(ans,c[i])+c[i]*(m>>(i-1)&1);
    }
    printf("%lld\n",ans);
    return 0;
}