#include "iostream"
#include "algorithm"
#include "cstring"
#include "cstdio"
#include "set"
#include "vector"
#include "map"
#include "cmath"
#define MAXN 506
//#define int long long
using namespace std;
typedef long long ll;
#define P 1000000007
int n;
int A[MAXN];

ll work( int l , int r , int L , int R ) {
    if( r - l <= 1 ) return 0;
    ll ans = 0x3f3f3f3f3f3f3f3f;
    for( int i = l + 1 ; i <= r - 1 ; ++ i ) ans = min( ans , work( l , i , L , L + R ) + work( i , r , L + R , R ) + 1ll * ( L + R ) * A[i] );
    return ans;
}

int main() {
    cin >> n;
    for( int i = 1 ; i <= n ; ++ i ) scanf("%d",&A[i]);
    cout << work( 1 , n , 1 , 1 ) + A[1] + A[n] << endl;
}
### 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
#include "iostream"
#include "algorithm"
#include "cstring"
#include "cstdio"
#include "set"
#include "vector"
#include "map"
#include "cmath"
#define MAXN 506
//#define int long long
using namespace std;
typedef long long ll;
#define P 1000000007
int n;
int A[MAXN];

ll work( int l , int r , int L , int R ) {
    if( r - l <= 1 ) return 0;
    ll ans = 0x3f3f3f3f3f3f3f3f;
    for( int i = l + 1 ; i <= r - 1 ; ++ i ) ans = min( ans , work( l , i , L , L + R ) + work( i , r , L + R , R ) + 1ll * ( L + R ) * A[i] );
    return ans;
}

int main() {
    cin >> n;
    for( int i = 1 ; i <= n ; ++ i ) scanf("%d",&A[i]);
    cout << work( 1 , n , 1 , 1 ) + A[1] + A[n] << endl;
}
```