#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N=19;
int n,a[N];
struct sta {
	int l,r,xl,xr;
};
bool operator < (sta A,sta B) {
	if(A.l != B.l) return A.l<B.l;
	if(A.r != B.r) return A.r<B.r;
	if(A.xl != B.xl) return A.xl<B.xl;
	return A.xr<B.xr;
}
map<sta,LL> Map;
LL solve(int l,int r,int xl,int xr) {
	if(l >= r-1) return 0;
	if(Map.count((sta){l,r,xl,xr})) return Map[(sta){l,r,xl,xr}];
	LL res=1e12;
	for(int i=l+1;i<r;++i)
		res=min(res,(LL)(xl+xr)*a[i]+solve(l,i,xl,xl+xr)+solve(i,r,xl+xr,xr));
	return Map[(sta){l,r,xl,xr}]=res;
}
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d",&a[i]);
	cout<<solve(1,n,1,1)+a[1]+a[n]<<endl;
	return 0;
}
### Prompt

In CPP, your task is to solve 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>
typedef long long LL;
using namespace std;
const int N=19;
int n,a[N];
struct sta {
	int l,r,xl,xr;
};
bool operator < (sta A,sta B) {
	if(A.l != B.l) return A.l<B.l;
	if(A.r != B.r) return A.r<B.r;
	if(A.xl != B.xl) return A.xl<B.xl;
	return A.xr<B.xr;
}
map<sta,LL> Map;
LL solve(int l,int r,int xl,int xr) {
	if(l >= r-1) return 0;
	if(Map.count((sta){l,r,xl,xr})) return Map[(sta){l,r,xl,xr}];
	LL res=1e12;
	for(int i=l+1;i<r;++i)
		res=min(res,(LL)(xl+xr)*a[i]+solve(l,i,xl,xl+xr)+solve(i,r,xl+xr,xr));
	return Map[(sta){l,r,xl,xr}]=res;
}
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d",&a[i]);
	cout<<solve(1,n,1,1)+a[1]+a[n]<<endl;
	return 0;
}
```