#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

LL s[25],n;

LL dp(LL l,LL r,LL sl,LL sr){
	LL ret = INF;
	if(l > r) return 0;
	for(LL i = l;i <= r;i ++){
		ret = min(ret,dp(l,i - 1,sl,sl + sr) + dp(i + 1,r,sl + sr,sr) + s[i] * (sl + sr));
	}
	return ret;
}

int main(){
	cin >> n;
	for(LL i = 1;i <= n;i ++) cin >> s[i];
	cout << dp(2,n - 1,1,1) + s[1] + s[n] << '\n';
	return 0;
}
### Prompt

Please formulate 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

LL s[25],n;

LL dp(LL l,LL r,LL sl,LL sr){
	LL ret = INF;
	if(l > r) return 0;
	for(LL i = l;i <= r;i ++){
		ret = min(ret,dp(l,i - 1,sl,sl + sr) + dp(i + 1,r,sl + sr,sr) + s[i] * (sl + sr));
	}
	return ret;
}

int main(){
	cin >> n;
	for(LL i = 1;i <= n;i ++) cin >> s[i];
	cout << dp(2,n - 1,1,1) + s[1] + s[n] << '\n';
	return 0;
}
```