#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int n, A[18];
map<pair<ll, ll>, ll> mem[18][18];

ll dfs(int l, int r, ll x, ll y) {
	if (r - l == 1)
		return 0;
	if (mem[l][r].count({x, y})) return mem[l][r][{x, y}];
	ll res = 1ll << 60;
	for (int i = l + 1; i < r; ++i)
		res = min(res, dfs(l, i, x, x + y) + dfs(i, r, x + y, y) + (x + y) * A[i]);
	return mem[l][r][{x, y}] = res;
}

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

### Prompt

Please create a solution in cpp 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;

typedef long long ll;

int n, A[18];
map<pair<ll, ll>, ll> mem[18][18];

ll dfs(int l, int r, ll x, ll y) {
	if (r - l == 1)
		return 0;
	if (mem[l][r].count({x, y})) return mem[l][r][{x, y}];
	ll res = 1ll << 60;
	for (int i = l + 1; i < r; ++i)
		res = min(res, dfs(l, i, x, x + y) + dfs(i, r, x + y, y) + (x + y) * A[i]);
	return mem[l][r][{x, y}] = res;
}

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

```