#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
typedef long long i64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 18+5;
int n; i64 a[N];
i64 dfs(int l, int r, int x, int y)
{
    if(l+1 == r) return 0; i64 ret = 1e18;
    for(int i = l+1; i < r; ++i) ret = min(ret, dfs(l, i, x, x+y)+dfs(i, r, x+y, y)+a[i]*(x+y));
    return ret;
}
int main()
{
    n = read(); for(int i = 1; i <= n; ++i) a[i] = read();
    printf("%lld\n", dfs(1, n, 1, 1)+a[1]+a[n]);
    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
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
typedef long long i64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 18+5;
int n; i64 a[N];
i64 dfs(int l, int r, int x, int y)
{
    if(l+1 == r) return 0; i64 ret = 1e18;
    for(int i = l+1; i < r; ++i) ret = min(ret, dfs(l, i, x, x+y)+dfs(i, r, x+y, y)+a[i]*(x+y));
    return ret;
}
int main()
{
    n = read(); for(int i = 1; i <= n; ++i) a[i] = read();
    printf("%lld\n", dfs(1, n, 1, 1)+a[1]+a[n]);
    return 0;
}
```