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 c

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：

按照店鋪單位價值排序一下，然後貪心加動態規劃

程式碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;
int n;
struct TA_
{
	ll v,w;
	TA_() {}
	TA_(ll a,ll b):v(a),w(b) {}
	bool operator< (const TA_& c)
	{
		//return (double)v/w > (double)c.v/c.w;  //單位價值大的在前 
		return v * c.w > c.v * w;   //交叉相乘
	}
} ta[35]; //店 

ll solve(ll left,ll cost,int i)//剩餘要買、當前花費、當前店鋪 
{
	ll m=left/ta[i].v; //要買的桶數，儘可能多的在當前店鋪買 
	cost+=m*ta[i].w;   //更新花費 
	left-=m*ta[i].v;   //更新要買 
	if(left==0) return cost; //如果剛好買完 
	if(i==n) return cost+ta[i].w; //如果還有要買且已經在最後一間店 
	ll temp=solve(left,cost,i+1); //去下一間店看看 
	return min(temp,cost+ta[i].w); //在當前店再買一桶，或者去下一間店買剩餘的 
}

int main(void)
{
	ll l;
	scanf("%d %lld",&n,&l);
	scanf("%lld",&ta[1].w);
	ta[1].v=1;
	for(int i=2;i<=n;i++) {
		scanf("%lld",&ta[i].w); 
		ta[i].v=ta[i-1].v<<1; 
	}
	sort(ta+1,ta+n+1);  
	printf("%lld\n",solve(l,0,1));
	return 0;
 }