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

void chmin(int64_t& a, int64_t b){
    a = min(a, b);
}

int N;
int64_t A[18];

map<vector<int>, int64_t> memo;

int64_t rec(int i, int j, int x, int y){
    if(j-i == 1) return 0;
    vector<int> v = {i, j, x, y};
    if(memo.count(v)) return memo[v];

    int64_t ret = 1e18;
    for(int m=i+1; m<j; m++) chmin(ret, rec(i, m, x, x+y) + rec(m, j, x+y, y) + (x+y)*A[m]);
    memo[v] = ret;
    return ret;
}

int main(){
    cin >> N;
    for(int i=0; i<N; i++) cin >> A[i];

    cout << A[0] + rec(0, N-1, 1, 1) + A[N-1] << endl;
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

void chmin(int64_t& a, int64_t b){
    a = min(a, b);
}

int N;
int64_t A[18];

map<vector<int>, int64_t> memo;

int64_t rec(int i, int j, int x, int y){
    if(j-i == 1) return 0;
    vector<int> v = {i, j, x, y};
    if(memo.count(v)) return memo[v];

    int64_t ret = 1e18;
    for(int m=i+1; m<j; m++) chmin(ret, rec(i, m, x, x+y) + rec(m, j, x+y, y) + (x+y)*A[m]);
    memo[v] = ret;
    return ret;
}

int main(){
    cin >> N;
    for(int i=0; i<N; i++) cin >> A[i];

    cout << A[0] + rec(0, N-1, 1, 1) + A[N-1] << endl;
    return 0;
}

```