#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,a[1010],cnt;
//map<ll,ll>f[18][18][280010];
ll solve(ll l,ll r,ll tl,ll tr){
	if(l+1==r)return 0;
	if(l+2==r)return (tl+tr)*a[l+1];
	cnt++;
//	if(f[l][r][tl][tr])return f[l][r][tl][tr];
	ll i,sej=1e18;
	for(i=l+1;i<r;i++)sej=min(sej,solve(l,i,tl,tl+tr)+solve(i,r,tl+tr,tr)+(tl+tr)*a[i]);
	return sej;
}
int main(){
	ll i;
	scanf("%lld",&n);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	//printf("%lld\n",cnt);
	printf("%lld\n",solve(1,n,1,1)+a[1]+a[n]);
	//printf("%lld",cnt);
}
/*
18
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
*/
### Prompt

Generate a Cpp solution to the following problem:

There is a stack of N cards, each of which has a non-negative integer written on it. The integer written on the i-th card from the top is A_i.

Snuke will repeat the following operation until two cards remain:

* Choose three consecutive cards from the stack.
* Eat the middle card of the three.
* For each of the other two cards, replace the integer written on it by the sum of that integer and the integer written on the card eaten.
* Return the two cards to the original position in the stack, without swapping them.



Find the minimum possible sum of the integers written on the last two cards remaining.

Constraints

* 2 \leq N \leq 18
* 0 \leq A_i \leq 10^9 (1\leq i\leq N)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the minimum possible sum of the integers written on the last two cards remaining.

Examples

Input

4
3 1 4 2


Output

16


Input

6
5 2 4 1 6 9


Output

51


Input

10
3 1 4 1 5 9 2 6 5 3


Output

115

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,a[1010],cnt;
//map<ll,ll>f[18][18][280010];
ll solve(ll l,ll r,ll tl,ll tr){
	if(l+1==r)return 0;
	if(l+2==r)return (tl+tr)*a[l+1];
	cnt++;
//	if(f[l][r][tl][tr])return f[l][r][tl][tr];
	ll i,sej=1e18;
	for(i=l+1;i<r;i++)sej=min(sej,solve(l,i,tl,tl+tr)+solve(i,r,tl+tr,tr)+(tl+tr)*a[i]);
	return sej;
}
int main(){
	ll i;
	scanf("%lld",&n);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	//printf("%lld\n",cnt);
	printf("%lld\n",solve(1,n,1,1)+a[1]+a[n]);
	//printf("%lld",cnt);
}
/*
18
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
*/
```