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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(),(x).end()
#define REP(i,n) for(int i=0;i<(n);++i)
typedef long long ll;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N; cin >> N;
    vector<ll> A(N);
    REP(i,N) cin >> A[i];
    sort(all(A));
    int idx = lower_bound(all(A),0LL) - A.begin();
    idx = max(1, idx);
    idx = min(N-1, idx);
    ll ans = A[N-1] - A[0];
    for(int i=1; i<N-1; ++i) {
        if(A[i]<0) ans -= A[i];
        else ans += A[i];
    }
    cout << ans << endl;
    ll t1 = A[N-1];
    for(int i=1; i<idx; ++i) {
        cout << t1 << " " << A[i] << '\n';
        t1 -= A[i];
    }
    ll t2 = A[0];
    for(int i=idx; i<N-1; ++i) {
        cout << t2 << " " << A[i] << '\n';
        t2 -= A[i];
    }
    cout << t1 << " " << t2 << '\n';
    return 0;
}