#include<bits/stdc++.h>
const int maxn = 18;
typedef long long ll;
ll dp[maxn][maxn][1 << maxn];
int a[maxn],n;
inline ll dfs(int l,int r,int lv,int rv,int st){
	if(~dp[l][r][st]) return dp[l][r][st];
	if(l + 1 == r) return (ll) lv * a[l] + (ll) rv * a[r];
	ll & ans = dp[l][r][st] = 1e18;
	for(int i = l + 1;i < r;++i){
		ans = std::min(ans,dfs(l,i,lv,lv + rv,st << 1) + dfs(i,r,lv + rv,rv,st << 1 | 1) - a[i] * ll(lv + rv));
	}
	return ans;
}
int main(){
	std::cin >> n, memset(dp,-1,sizeof dp);
	for(int i = 0;i < n;++i) std::cin >> a[i];
	std::cout << dfs(0,n - 1,1,1,1) << '\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<bits/stdc++.h>
const int maxn = 18;
typedef long long ll;
ll dp[maxn][maxn][1 << maxn];
int a[maxn],n;
inline ll dfs(int l,int r,int lv,int rv,int st){
	if(~dp[l][r][st]) return dp[l][r][st];
	if(l + 1 == r) return (ll) lv * a[l] + (ll) rv * a[r];
	ll & ans = dp[l][r][st] = 1e18;
	for(int i = l + 1;i < r;++i){
		ans = std::min(ans,dfs(l,i,lv,lv + rv,st << 1) + dfs(i,r,lv + rv,rv,st << 1 | 1) - a[i] * ll(lv + rv));
	}
	return ans;
}
int main(){
	std::cin >> n, memset(dp,-1,sizeof dp);
	for(int i = 0;i < n;++i) std::cin >> a[i];
	std::cout << dfs(0,n - 1,1,1,1) << '\n';
}

```