#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
using vi = vector<ll>;
using pi = pair<ll, ll>;
const int maxn = 18;//, mod = 1e9 + 7;
int n, a[maxn];
map<pi, ll> dp[maxn][maxn];
ll f(ll l, ll r, ll xl, ll xr) {
	if(l>r) return 0;
	if(dp[l][r].count({xl, xr})) return dp[l][r][{xl, xr}];
	ll ans = 1ll<<60;
	for(int i = l; i <= r; i++) {
		ans = min(ans, f(l, i-1, xl, xl+xr) + f(i+1, r, xl+xr, xr) + (xl+xr)*a[i]);
	}
	return dp[l][r][{xl, xr}] = ans;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	cout << f(1, n-2, 1, 1)+a[0]+a[n-1];
}


### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
using namespace std;
using ll = long long;
using vi = vector<ll>;
using pi = pair<ll, ll>;
const int maxn = 18;//, mod = 1e9 + 7;
int n, a[maxn];
map<pi, ll> dp[maxn][maxn];
ll f(ll l, ll r, ll xl, ll xr) {
	if(l>r) return 0;
	if(dp[l][r].count({xl, xr})) return dp[l][r][{xl, xr}];
	ll ans = 1ll<<60;
	for(int i = l; i <= r; i++) {
		ans = min(ans, f(l, i-1, xl, xl+xr) + f(i+1, r, xl+xr, xr) + (xl+xr)*a[i]);
	}
	return dp[l][r][{xl, xr}] = ans;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i];
	cout << f(1, n-2, 1, 1)+a[0]+a[n-1];
}


```