#include <bits/stdc++.h>

using namespace std;

const int MX = 18;

int a[MX];

map<tuple<int, int, int, int>, long long> dp;

long long solve(int l, int r, int lv, int rv) {
	if (l > r) return 0;
	
	auto t = make_tuple(l, r, lv, rv);
	if (dp.count(t) == 0) {
		long long best = 1e18, cv = lv + rv;
		for (int i = l; i <= r; i++) {
			best = min(solve(l, i - 1, lv, cv) + solve(i + 1, r, cv, rv) + a[i] * cv, best);
		}
		
		dp[t] = best;
	}
	
	return dp[t];
}

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
	
	auto ans = solve(1, n - 2, 1, 1) + a[0] + a[n - 1];
	
	printf("%lld\n", ans);
	
	return 0;
}

### Prompt

Please formulate 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
#include <bits/stdc++.h>

using namespace std;

const int MX = 18;

int a[MX];

map<tuple<int, int, int, int>, long long> dp;

long long solve(int l, int r, int lv, int rv) {
	if (l > r) return 0;
	
	auto t = make_tuple(l, r, lv, rv);
	if (dp.count(t) == 0) {
		long long best = 1e18, cv = lv + rv;
		for (int i = l; i <= r; i++) {
			best = min(solve(l, i - 1, lv, cv) + solve(i + 1, r, cv, rv) + a[i] * cv, best);
		}
		
		dp[t] = best;
	}
	
	return dp[t];
}

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
	
	auto ans = solve(1, n - 2, 1, 1) + a[0] + a[n - 1];
	
	printf("%lld\n", ans);
	
	return 0;
}

```