// ※※※ 解答不能 ※※※
// 正解者のソースを学習する(一部改変).
// tourist氏
// https://atcoder.jp/contests/agc035/submissions/6377009
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
vector<LL> a;
int N;

LL Go(int from, int to, LL ca, LL cb){
    if(to - from == 1) return 0;
    LL best = 1e18;
    for(int i = from + 1; i < to; i++){
        LL cur = a[i] * (ca + cb) + Go(from, i, ca, ca + cb) + Go(i, to, ca + cb, cb);
        best = min(best, cur);
    }
    return best;
}

int main() {
    scanf("%d", &N);
    a.resize(N);
    for(int i = 0; i < N; i++) scanf("%d", &a[i]);
    printf("%lld\n", Go(0, N - 1, 1, 1) + a[0] + a[N - 1]);
    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
// ※※※ 解答不能 ※※※
// 正解者のソースを学習する(一部改変).
// tourist氏
// https://atcoder.jp/contests/agc035/submissions/6377009
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
vector<LL> a;
int N;

LL Go(int from, int to, LL ca, LL cb){
    if(to - from == 1) return 0;
    LL best = 1e18;
    for(int i = from + 1; i < to; i++){
        LL cur = a[i] * (ca + cb) + Go(from, i, ca, ca + cb) + Go(i, to, ca + cb, cb);
        best = min(best, cur);
    }
    return best;
}

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