// D.cpp
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MAX_N = 20, INF = 0x3f3f3f3f3f3f3f3f;

int n;
ll ai[MAX_N];

ll solve(int l, int r, ll cL, ll cR)
{
    if (l + 1 > r - 1)
        return 0;
    ll ret = INF;
    for (int i = l + 1; i <= r - 1; i++)
        ret = min(ret, solve(l, i, cL + cR, cR) + solve(i, r, cL, cL + cR) + 1LL * (cL + cR) * ai[i]);
    return ret;
}

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

Develop a solution in cpp to the problem described below:

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
// D.cpp
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MAX_N = 20, INF = 0x3f3f3f3f3f3f3f3f;

int n;
ll ai[MAX_N];

ll solve(int l, int r, ll cL, ll cR)
{
    if (l + 1 > r - 1)
        return 0;
    ll ret = INF;
    for (int i = l + 1; i <= r - 1; i++)
        ret = min(ret, solve(l, i, cL + cR, cR) + solve(i, r, cL, cL + cR) + 1LL * (cL + cR) * ai[i]);
    return ret;
}

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