/*
 * Author       : YangDavid
 * Created Time : 2019年07月15日 星期一 11时24分59秒
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const ll INFL = 0x3f3f3f3f3f3f3f3fLL;
int n, a[500];
void upd(ll& x, ll y) { x = min(x, y); }

ll solve(int l, int r, ll wl, ll wr) {
    if(l > r) return 0;
    ll ret = INFL;
    for(int i = l; i <= r; ++i)
        upd(ret, solve(l,i-1,wl,wr+wl)+solve(i+1, r,wl+wr,wr)+(wl+wr)*a[i]);
    return ret;
}
int main() {
    scanf("%d", &n);
    rep(i, n) scanf("%d", &a[i]);
    printf("%lld\n", solve(2, n - 1, 1, 1) + a[1] + a[n]);

    return 0;
}

### Prompt

Develop a solution in CPP 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
/*
 * Author       : YangDavid
 * Created Time : 2019年07月15日 星期一 11时24分59秒
 */

#include<bits/stdc++.h>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const ll INFL = 0x3f3f3f3f3f3f3f3fLL;
int n, a[500];
void upd(ll& x, ll y) { x = min(x, y); }

ll solve(int l, int r, ll wl, ll wr) {
    if(l > r) return 0;
    ll ret = INFL;
    for(int i = l; i <= r; ++i)
        upd(ret, solve(l,i-1,wl,wr+wl)+solve(i+1, r,wl+wr,wr)+(wl+wr)*a[i]);
    return ret;
}
int main() {
    scanf("%d", &n);
    rep(i, n) scanf("%d", &a[i]);
    printf("%lld\n", solve(2, n - 1, 1, 1) + a[1] + a[n]);

    return 0;
}

```