#include <bits/stdc++.h>

#define R register
#define ll long long
#define sum(a, b, mod) (((a) + (b)) % mod)

ll n, a[50];

ll f(ll l, ll r, ll x, ll y)
{
    ll ans = 1e18;
    if(r - l <= 1) return 0;
    for(int i = l + 1; i <= r - 1; i++)
        ans = std::min((f(l, i, x, x + y) + f(i, r, x + y, y) + a[i] * (x + y)), ans);
    return ans;
}

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

### Prompt

Your challenge is to write 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>

#define R register
#define ll long long
#define sum(a, b, mod) (((a) + (b)) % mod)

ll n, a[50];

ll f(ll l, ll r, ll x, ll y)
{
    ll ans = 1e18;
    if(r - l <= 1) return 0;
    for(int i = l + 1; i <= r - 1; i++)
        ans = std::min((f(l, i, x, x + y) + f(i, r, x + y, y) + a[i] * (x + y)), ans);
    return ans;
}

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

```