Problem:
Score : 500 points

Problem Statement

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}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(false), cout << fixed << setprecision(20);
#define int long long
#define gcd __gcd
#define all(x) (x).begin(), (x).end()
template<class T>
bool chmax(T& a, T b){return (a = max(a, b)) == b;}
template<class T>
bool chmin(T& a, T b){return (a = min(a, b)) == b;}
#define _overload(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(...) _overload(__VA_ARGS__, repi, _rep)(__VA_ARGS__)
#define _rev(i, n) revi(i, n, 0)
#define revi(i, a, b) for(int i = (int)(a - 1); i >= (int)(b); i--)
#define rev(...) _overload(__VA_ARGS__, revi, _rev)(__VA_ARGS__)
#define each(i, n) for(auto&& i: n)
const int INF = 1e18, MOD = 1e9 + 7;
signed main() {
    INCANT;
    int n, a, mn, mx;
    deque<int> q;
    vector<pair<int, int>> res;
    cin>>n;
    rep(i, n) {
        cin>>a;
        q.push_back(a);
    }
    sort(all(q));
    mx = q.back();
    mn = q.front();
    q.pop_back();
    q.pop_front();
    while(q.front() < 0) {
        res.push_back({mx, q.front()});
        mx -= q.front();
        q.pop_front();
    }
    while(!q.empty()) {
        res.push_back({mn, q.back()});
        mn -= q.back();
        q.pop_back();
    }
    cout<<mx - mn<<endl;
    each(p, res) {
        cout<<p.first<<' '<<p.second<<endl;
    }
    cout<<mx<<' '<<mn<<endl;
}