#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, pii> piipii;
map<piipii, ll> dp;
int a[20];
 
ll dfs(int l, int r, int al, int ar){
    if(r-l+1 <= 2) return 0;
    if(dp.count(piipii(pii(l, r), pii(al, ar)))) return dp[piipii(pii(l, r), pii(al, ar))];
    ll res = 1e18;
    for(int i=l+1;i<=r-1;i++){
        res = min(res, a[i]*1ll*(al+ar) + dfs(l, i, al, al+ar) + dfs(i, r, al+ar, ar));
    }
    return dp[piipii(pii(l, r), pii(al, ar))] = res;
}
 
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]);
    printf("%lld\n", dfs(1, n, 1, 1) + a[1] + a[n]);
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, pii> piipii;
map<piipii, ll> dp;
int a[20];
 
ll dfs(int l, int r, int al, int ar){
    if(r-l+1 <= 2) return 0;
    if(dp.count(piipii(pii(l, r), pii(al, ar)))) return dp[piipii(pii(l, r), pii(al, ar))];
    ll res = 1e18;
    for(int i=l+1;i<=r-1;i++){
        res = min(res, a[i]*1ll*(al+ar) + dfs(l, i, al, al+ar) + dfs(i, r, al+ar, ar));
    }
    return dp[piipii(pii(l, r), pii(al, ar))] = res;
}
 
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]);
    printf("%lld\n", dfs(1, n, 1, 1) + a[1] + a[n]);
}
```