/*
	记录状态不太好搞
	考虑计算每个数被计算多少次
	分制 + 递归 
*/
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <ctime>
#define N 20
#define inf 1234567898765431LL

using namespace std;

typedef long long LL;

int n;
int a[N];

// 复杂度不会证，但好像比 n! 小很多 
LL solve(int l, int r, int x, int y) {
	if(l + 1 == r) return 0;
	LL res = inf;
	for(int i = l + 1; i < r; ++i) {
		LL re = 1LL * (x + y) * a[i] + 
			    solve(l, i, x, x + y) +
		 	    solve(i, r, x + y, y);
		if(re < res) res = re;
	}
	return res;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	printf("%lld\n", a[1] + solve(1, n, 1, 1) + a[n]);
	return 0;
}
### Prompt

Please create a solution in Cpp 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
/*
	记录状态不太好搞
	考虑计算每个数被计算多少次
	分制 + 递归 
*/
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <ctime>
#define N 20
#define inf 1234567898765431LL

using namespace std;

typedef long long LL;

int n;
int a[N];

// 复杂度不会证，但好像比 n! 小很多 
LL solve(int l, int r, int x, int y) {
	if(l + 1 == r) return 0;
	LL res = inf;
	for(int i = l + 1; i < r; ++i) {
		LL re = 1LL * (x + y) * a[i] + 
			    solve(l, i, x, x + y) +
		 	    solve(i, r, x + y, y);
		if(re < res) res = re;
	}
	return res;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	printf("%lld\n", a[1] + solve(1, n, 1, 1) + a[n]);
	return 0;
}
```