// TODO: Why there are at most O(2^n poly(n)) states?
// Solution by official editorial

#include<cstdio>

using ll = long long;

constexpr ll inf = 0x3f3f3f3f3f3f3f3f;

inline void dmin(ll &m, ll n) {
	if (n < m) m = n;
}

ll z[21];

ll dfs(int l, int r, int xl, int xr) {
	if (l + 1 == r) return z[l] * xl + z[r] * xr;
	ll ans = inf;
	for (int m = l + 1; m < r; ++m)
		dmin(ans, dfs(l, m, xl, xl + xr) + dfs(m, r, xl + xr, xr) - z[m] * (xl + xr));
	return ans;
}

int main() {
	int N;

	scanf("%d", &N);
	for (int i = 1; i <= N; ++i)
		scanf("%lld", z + i);

	printf("%lld", dfs(1, N, 1, 1));
	return 0;
}


### Prompt

Generate a cpp solution 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
// TODO: Why there are at most O(2^n poly(n)) states?
// Solution by official editorial

#include<cstdio>

using ll = long long;

constexpr ll inf = 0x3f3f3f3f3f3f3f3f;

inline void dmin(ll &m, ll n) {
	if (n < m) m = n;
}

ll z[21];

ll dfs(int l, int r, int xl, int xr) {
	if (l + 1 == r) return z[l] * xl + z[r] * xr;
	ll ans = inf;
	for (int m = l + 1; m < r; ++m)
		dmin(ans, dfs(l, m, xl, xl + xr) + dfs(m, r, xl + xr, xr) - z[m] * (xl + xr));
	return ans;
}

int main() {
	int N;

	scanf("%d", &N);
	for (int i = 1; i <= N; ++i)
		scanf("%lld", z + i);

	printf("%lld", dfs(1, N, 1, 1));
	return 0;
}


```