#include <bits/stdc++.h>

using namespace std;
using ll=long long;
int a[20];

ll dp(int l, int r, ll x, ll y) {
    if (l > r) return 0;
    if (l == r) return (x+y) * a[l];
    ll mi = 1ll<<62;
    for (int m = l; m <= r; m++) {
        mi = min(mi, dp(l, m-1, x, x+y) + dp(m+1, r, x+y, y) + (x+y)*a[m]);
    }
    return mi;
}
void solve() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cout << dp(1, n-2, 1, 1) + a[0] + a[n-1];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

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

using namespace std;
using ll=long long;
int a[20];

ll dp(int l, int r, ll x, ll y) {
    if (l > r) return 0;
    if (l == r) return (x+y) * a[l];
    ll mi = 1ll<<62;
    for (int m = l; m <= r; m++) {
        mi = min(mi, dp(l, m-1, x, x+y) + dp(m+1, r, x+y, y) + (x+y)*a[m]);
    }
    return mi;
}
void solve() {
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    cout << dp(1, n-2, 1, 1) + a[0] + a[n-1];
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}

```