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.

The question ： You need to buy water , At least l l , Shui you n Species specification , The second line n A digital , It means the first one i Growing water requires ci element , The first i The weight of the seed water is 2^(i-1) l , Please spend at least x Yuan at least l Premium .

Ideas ： Suppose you can only buy l Premium , You can't buy more or less . Deal with the data , Will buy 1,2,4... How much does it cost to raise water at least ,（if(ci>2*c(i-1)) that ,ci=2*ci-1) Then the weight of water to be bought can be represented by a binary string of water to be sold .（ here x+=((l>>i)&1)*ci;)

You can buy more because of this question , That is, the value of the last binary string will be greater than or equal to the value of the previous binary string ,( for example 13(1101), Can buy 14(1110)) At this time, just add a judgment condition to the previous formula （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;
}