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: Wrong Answer)

#include <stdio.h>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <numeric>
#include <complex>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <cassert>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long ll;
const double EPS = 1e-9;
typedef vector<int> vint;
typedef vector<vector<int>> v2int;
typedef vector<ll> vll;
typedef vector<vector<ll>> v2ll;
typedef list<int> liint;
typedef pair<int, int> pint;
const int INF = int(2e9);
const ll LINF = ll(2e9) * ll(2e9);
#define rep(i, n) REP(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define MSG(a) cout << #a << " " << a << endl;
#define REP(i, x, n) for(int i = x; i < n; i++)
template<class T, class C> void chmax(T& a, C b) { a > b ? : a = b; }
template<class T, class C> void chmin(T& a, C b) { a < b ? : a = b; }

int gcd(int a, int b) {
    int r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

int main() {
    int N;
    cin >> N;
    vint A(N);
    bool is_all_minus = true;
    bool is_all_plus = true;
    int first_minus_idx = -1;
    int first_plus_idx = -1;
    rep(i, N) {
        cin >> A[i];
        if (A[i] < 0) {
            is_all_plus = false;
            first_minus_idx = i;
        }
        if (A[i] >= 0) {
            is_all_minus = false;
            first_plus_idx = i;
        }
        if (A[i] == 0)
            first_plus_idx = i;
    }
    ll ans = 0;
    vector<pair<ll, ll>> ans_list(N);
    if (is_all_plus == true) {
        auto  minA_it = min_element(A.begin(), A.end());
        int minA_idx = distance(A.begin(), minA_it);
        int last_idx = 0;
        if (minA_idx == 0)
            last_idx = 1;
        ans = A[minA_idx];
        int j = 0;
        REP(i, 0, N) {
            if (i != minA_idx && i != last_idx) {
                ans_list[j] = make_pair(ans, A[i]);
                ans = ans - A[i];
                j++;
            }
        }
        ans_list[j] = make_pair(A[last_idx], ans);
        ans = A[last_idx] - ans;
    }
    else if (is_all_minus == true) {
        auto  maxA_it = max_element(A.begin(), A.end());
        int maxA_idx = distance(A.begin(), maxA_it);
        int last_idx = 0;
        if (maxA_idx == 0)
            last_idx = 1;
        ans = A[maxA_idx];
        int j = 0;
        REP(i, 0, N) {
            if (i != maxA_idx && i != last_idx) {
                ans_list[j] = make_pair(ans, A[i]);
                ans = ans - A[i];
                j++;
            }
        }
        ans_list[j] = make_pair(A[last_idx], ans);
        ans = A[last_idx] - ans;
    }
    else {
        ans = A[first_minus_idx];
        int j = 0;
        rep(i, N) {
            if (i != first_minus_idx && A[i] >= 0 && i != first_plus_idx) {
                ans_list[j] = make_pair(ans, A[i]);
                ans = ans - A[i];
                j++;
            }
        }
        ans_list[j] = make_pair(A[first_plus_idx], ans);
        ans = A[first_plus_idx] - ans;
        j++;
        rep(i, N) {
            if (A[i] < 0 && i != first_minus_idx) {
                ans_list[j] = make_pair(ans, A[i]);
                ans = ans - A[i];
                j++;
            }
        }
    }
    cout << ans << endl;
    rep(i, N - 1) {
        cout << ans_list[i].first << " " << ans_list[i].second << endl;
    }
    return 0;
}