#include<cstdio>
#define M 131072
#define N 20
inline long long min(long long x,long long y){return x<y?x:y;}
int e[M],n;long long f[M][N][N],g[M],h[M];bool c[M][N][N];
long long dfs(int u,int v,int w)
{
    if(v+1==w)return 0ll;
    if(c[u][v][w])return f[u][v][w];
    c[u][v][w]=true,f[u][v][w]=0x7fffffffffffffffll;
    for(int i=v+1;i<w;i++)f[u][v][w]=min(f[u][v][w],dfs(u<<1,v,i)+dfs(u<<1|1,i,w)+e[i]*(g[u]+h[u]));
    return f[u][v][w];
}
int main()
{
    scanf("%d",&n),g[1]=h[1]=1;
    for(int i=1;i<=n;i++)scanf("%d",e+i);
    for(int i=1;i<1<<n-2;i++)g[i<<1]=g[i],h[i<<1]=g[i<<1|1]=g[i]+h[i],h[i<<1|1]=h[i];
    return 0&(printf("%lld\n",e[1]+e[n]+dfs(1,1,n)));
}
### Prompt

Create a solution in CPP for 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>
#define M 131072
#define N 20
inline long long min(long long x,long long y){return x<y?x:y;}
int e[M],n;long long f[M][N][N],g[M],h[M];bool c[M][N][N];
long long dfs(int u,int v,int w)
{
    if(v+1==w)return 0ll;
    if(c[u][v][w])return f[u][v][w];
    c[u][v][w]=true,f[u][v][w]=0x7fffffffffffffffll;
    for(int i=v+1;i<w;i++)f[u][v][w]=min(f[u][v][w],dfs(u<<1,v,i)+dfs(u<<1|1,i,w)+e[i]*(g[u]+h[u]));
    return f[u][v][w];
}
int main()
{
    scanf("%d",&n),g[1]=h[1]=1;
    for(int i=1;i<=n;i++)scanf("%d",e+i);
    for(int i=1;i<1<<n-2;i++)g[i<<1]=g[i],h[i<<1]=g[i<<1|1]=g[i]+h[i],h[i<<1|1]=h[i];
    return 0&(printf("%lld\n",e[1]+e[n]+dfs(1,1,n)));
}
```