#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
typedef long long ll;
map<pii,ll> dp[20][20];ll num[22];
ll dfs(int l,int r,int lx,int rx)
{
    if(l>=r-1)return 0;
    if(dp[l][r].count({lx,rx}))return dp[l][r][{lx,rx}];
    dp[l][r][{lx,rx}]=1e18;
    rep(i,l+1,r)
    {
        ll res=num[i]*(lx+rx);
        res+=dfs(l,i,lx,lx+rx)+dfs(i,r,lx+rx,rx);
        dp[l][r][{lx,rx}]=min(dp[l][r][{lx,rx}],res);
    }
    return dp[l][r][{lx,rx}];
}
int main()
{
    int n;scanf("%d",&n);
    rep(i,1,n+1)scanf("%lld",&num[i]);
    printf("%lld\n",dfs(1,n,1,1)+num[1]+num[n]);
}
### Prompt

Generate 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;
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
typedef long long ll;
map<pii,ll> dp[20][20];ll num[22];
ll dfs(int l,int r,int lx,int rx)
{
    if(l>=r-1)return 0;
    if(dp[l][r].count({lx,rx}))return dp[l][r][{lx,rx}];
    dp[l][r][{lx,rx}]=1e18;
    rep(i,l+1,r)
    {
        ll res=num[i]*(lx+rx);
        res+=dfs(l,i,lx,lx+rx)+dfs(i,r,lx+rx,rx);
        dp[l][r][{lx,rx}]=min(dp[l][r][{lx,rx}],res);
    }
    return dp[l][r][{lx,rx}];
}
int main()
{
    int n;scanf("%d",&n);
    rep(i,1,n+1)scanf("%lld",&num[i]);
    printf("%lld\n",dfs(1,n,1,1)+num[1]+num[n]);
}
```