#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <climits>
#define N 18
using namespace std;

int a[N];

long long DP(int l, int r, int x, int y)
{
    int i;
    long long o;

    if(l + 1 == r)
        return (long long)a[l] * x + (long long)a[r] * y;

    for(i = l + 1, o = LLONG_MAX; i < r; i ++)
        o = min(o, DP(l, i, x, x + y) + DP(i, r, x + y, y) - (long long)a[i] * (x + y));

    return o;
}

int main(void)
{
    int n;
    int i;

    cin >> n;
    for(i = 0; i < n; i ++)
        cin >> a[i];
    cout << DP(0, n - 1, 1, 1) << endl;

    return 0;
}

### Prompt

Your task is to create 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
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <climits>
#define N 18
using namespace std;

int a[N];

long long DP(int l, int r, int x, int y)
{
    int i;
    long long o;

    if(l + 1 == r)
        return (long long)a[l] * x + (long long)a[r] * y;

    for(i = l + 1, o = LLONG_MAX; i < r; i ++)
        o = min(o, DP(l, i, x, x + y) + DP(i, r, x + y, y) - (long long)a[i] * (x + y));

    return o;
}

int main(void)
{
    int n;
    int i;

    cin >> n;
    for(i = 0; i < n; i ++)
        cin >> a[i];
    cout << DP(0, n - 1, 1, 1) << endl;

    return 0;
}

```