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++ (GCC 9.2.1) (judge verdict: Accepted)

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

// 変数ダンプ先。coutかcerr
#define DUMPOUT cerr

// 提出時はコメントアウト
#define DEBUG_

// #define int long long // intで書いたけど心配なときにlong longに変換する
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);}} fast;

/* cpp template {{{ */

/* short */
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) begin(v), end(v)
#define RALL(v) rbegin(v), rend(v)
#define X real()
#define Y imag()

/* REPmacro */
#define REPS(i, a, n) for (ll i = (a); i < (ll)(n); ++i)
#define REP(i, n) REPS(i, 0, n)
#define RREP(i, n) REPS(i, 1, n + 1)
#define DEPS(i, a, n) for (ll i = (a); i >= (ll)(n); --i)
#define DEP(i, n) DEPS(i, n, 0)
#define EACH(i, n) for (auto&& i : n)

/* debug */
#define debug(x) cerr << x << " " << "(L:" << __LINE__ << ")" << '\n';

/* alias */
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using D = double;
using P = complex<D>;
using vs = vector<string>;
template <typename T> using PQ = priority_queue<T>;
template <typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7;
const D EPS = 1e-9;
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

/* func */
inline bool inside(int y, int x, int H, int W) {return y >= 0 && x >= 0 && y < H && x < W;}
inline int in() {int x; cin >> x; return x;}
inline ll IN() {ll x; cin >> x; return x;}
inline vs split(const string& t, char c) {vs v; stringstream s(t); string b; while(getline(s, b, c)) v.eb(b); return v;}
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
template <typename T, typename S> inline void print(const pair<T, S>& p) {cout << p.first << " " << p.second << endl;}
template <typename T> inline void print(const T& x) {cout << x << '\n';}
template <typename T, typename S> inline void print(const vector<pair<T, S>>& v) {for (auto&& p : v) print(p);}
template <typename T> inline void print(const vector<T>& v, string s = " ") {REP(i, v.size()) cout << v[i] << (i != (ll)v.size() - 1 ? s : "\n");}
// clang-format on
/* }}} */

#ifdef DEBUG_
#define DEB
#else
// DEB と打つとデバッグ時以外はコメントアウトになる
#define DEB /##/
#endif

// 変数ダンプ用マクロ。デバッグ時以外は消滅する
// 引数はいくつでもどんな型でも可（ストリーム出力演算子があればOK）
#define dump(...) DEB DUMPOUT<<"  "; \
DUMPOUT<<#__VA_ARGS__<<" :["<<__LINE__<<":"<<__FUNCTION__<<"]"<<endl; \
DUMPOUT<<"    "; \
dump_func(__VA_ARGS__)

// デバッグ用変数ダンプ関数
void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail)
{
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) {
        DUMPOUT << " ";
    }
    else {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}

// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
    os << "{";
    for (int i = 0; i<vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}



signed main() {
    ll N; cin >> N;
    vector<ll> a(N); REP(i, N) cin >> a[i];
    sort(ALL(a));

    vector<ll> am;
    vector<ll> azp;
    vector<pair<ll, ll>> procs;

    REP(i, N) {
        if (a[i] < 0) am.pb(a[i]);
        else azp.pb(a[i]);
    }

    sort(ALL(am));
    sort(ALL(azp), greater<ll>());

    if (am.size() == N) {
        azp.pb(am.back());
        am.pop_back();
    }
    if (azp.size() == N) {
        am.pb(azp.back());
        azp.pop_back();
    }

    ll x, y, num;
    while (azp.size() > 1) {
        x = am.back();
        y = azp.back();
        procs.pb(mp(x, y));
        num = x - y;
        am.pop_back();
        azp.pop_back();
        am.pb(num);
    }

    while (am.size() > 0) {
        x = azp.back();
        y = am.back();
        procs.pb(mp(x, y));
        num = x - y;
        am.pop_back();
        azp.pop_back();
        azp.pb(num);
    }

    ll ans = azp[0];
    cout << ans << endl;

    REP(i, procs.size()) {
        cout << procs[i].Fi << ' ' << procs[i].Se << endl;
    }

    return 0;
}

// https://github.com/kurokoji/.cpp-Template/wiki テンプレートについて
// http://www.creativ.xyz/dump-cpp-652 dump()について
// https://gist.github.com/rigibun/7905920 色々