#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

map <ll, ll> dp[18][18];
ll A[22];
ll n;

ll f(ll l, ll r, ll x, ll y)
{
	if(l > r) return 0;
	if(l == r) return A[l] * x + A[l] * y;
	if(dp[l][r][x * 10000 + y]) return dp[l][r][x * 10000 + y];
	
	ll ret = 1e18, i;
	
	for(i=l; i<=r; i++){
		ret = min(ret, f(l, i - 1, x, x + y) + f(i + 1, r, x + y, y) + A[i] * (x + y));
	}
	
	return dp[l][r][x * 10000 + y] = ret;
}


int main()
{
	ll i;
	
	scanf("%lld", &n);
	
	for(i=0; i<n; i++){
		scanf("%lld", A + i);
	}
	
	printf("%lld\n", f(1, n - 2, 1, 1) + A[0] + A[n - 1]);
	
	return 0;
}

### 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>

using namespace std;

typedef long long ll;

map <ll, ll> dp[18][18];
ll A[22];
ll n;

ll f(ll l, ll r, ll x, ll y)
{
	if(l > r) return 0;
	if(l == r) return A[l] * x + A[l] * y;
	if(dp[l][r][x * 10000 + y]) return dp[l][r][x * 10000 + y];
	
	ll ret = 1e18, i;
	
	for(i=l; i<=r; i++){
		ret = min(ret, f(l, i - 1, x, x + y) + f(i + 1, r, x + y, y) + A[i] * (x + y));
	}
	
	return dp[l][r][x * 10000 + y] = ret;
}


int main()
{
	ll i;
	
	scanf("%lld", &n);
	
	for(i=0; i<n; i++){
		scanf("%lld", A + i);
	}
	
	printf("%lld\n", f(1, n - 2, 1, 1) + A[0] + A[n - 1]);
	
	return 0;
}

```