There are N integers, A_1, A_2, ..., A_N, written on a blackboard.

We will repeat the following operation N-1 times so that we have only one integer on the blackboard.

* Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y.



Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.

Constraints

* 2 \leq N \leq 10^5
* -10^4 \leq A_i \leq 10^4
* 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 maximum possible value M of the final integer on the blackboard, and a sequence of operations x_i, y_i that maximizes the final integer, in the format below.

Here x_i and y_i represent the integers x and y chosen in the i-th operation, respectively.

If there are multiple sequences of operations that maximize the final integer, any of them will be accepted.


M
x_1 y_1
:
x_{N-1} y_{N-1}

Output

Print the maximum possible value M of the final integer on the blackboard, and a sequence of operations x_i, y_i that maximizes the final integer, in the format below.

Here x_i and y_i represent the integers x and y chosen in the i-th operation, respectively.

If there are multiple sequences of operations that maximize the final integer, any of them will be accepted.


M
x_1 y_1
:
x_{N-1} y_{N-1}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include <bits/stdc++.h>

using namespace std;

#define int long long

signed main()
{
    cin.tie(0), cout.tie(0), ios::sync_with_stdio(0);

    int n;
    cin >> n;
    vector<int> vec(n);
    for (int i = 0; i < n; i++)
        cin >> vec[i];
    sort(vec.begin(), vec.end());
    vector<pair<int, int>> log;
    int ans = vec[0];
    int cur = n - 2;
    while (cur > 0 && vec[cur] >= 0)
    {
        log.push_back({ans, vec[cur]});
        ans -= vec[cur];
        cur--;
    }

    log.push_back({vec.back(), ans});
    ans = vec.back() - ans;
    while (cur > 0)
    {
        log.push_back({ans, vec[cur]});
        ans -= vec[cur];
        cur--;
    }

    cout << ans << endl;
    for (auto i: log)
        cout << i.first << " " << i.second << endl;

    return 0;
}
