#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;

int n;
ll a[18];

ll solve (int l, int r, int xl, int xr) {
    if (l+1 == r) return 0;
    ll ret = linf;
    for (int i = l+1; i < r; i++) {
        ret = min(ret, solve(l, i, xl, xl+xr) + solve(i, r, xl+xr, xr) + a[i] * (xl+xr));
    }
    return ret;
}

int main() {
    cin >> n;
    rep(i,n) cin >> a[i];
    cout << solve(0, n-1, 1, 1) + a[0] + a[n-1] << endl;
}

### Prompt

Please provide a CPP coded solution 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
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;

int n;
ll a[18];

ll solve (int l, int r, int xl, int xr) {
    if (l+1 == r) return 0;
    ll ret = linf;
    for (int i = l+1; i < r; i++) {
        ret = min(ret, solve(l, i, xl, xl+xr) + solve(i, r, xl+xr, xr) + a[i] * (xl+xr));
    }
    return ret;
}

int main() {
    cin >> n;
    rep(i,n) cin >> a[i];
    cout << solve(0, n-1, 1, 1) + a[0] + a[n-1] << endl;
}

```