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 "bits/stdc++.h"
using namespace std;    

#define COUNTOF(array) (sizeof(array) / sizeof(array[0]))    
#define OUT(P) cout << (P) << endl    
#define OUT2(P1, P2) cout << (P1) << " " << (P2) << endl    
#define PRINT_ARRAY(P) for(int i=0; i<P.size(); i++) cout << P[i] << endl;    
#define PRINT_ARRAY_INLINE(P) do { for(int i=0; i<P.size(); i++) cout << P[i] << " "; cout << endl; } while(0)    
#define PRINT_ARRAY_WITH_LABEL(L, P) do { cout << L << ": "; PRINT_ARRAY_INLINE(P); cout << endl; } while(0)    
typedef long long ll;    

int sum_digits(int num);    
template <class T>    
vector<T> copy_vector(vector<T> vec);    
int gcd(int a, int b);    

int main() {    
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i=0; i<n; i++) {
        cin >> a[i];
    }
    sort(a.rbegin(), a.rend());

    stack<int> st;
    for(int i=0; i<n; i++) {
        st.push(a[i]);
    }

    vector<pair<int, int>> steps;
    while(st.size() > 2) {
        int x1 = st.top();
        st.pop();
        int x2 = st.top();
        st.pop();
        int y = x1 - x2; 
        steps.push_back(make_pair(x1, x2));
        st.push(y);
    }

    int x1 = st.top();
    st.pop();
    int x2 = st.top();
    st.pop();
    int ans = x2 - x1;
    steps.push_back(make_pair(x2, x1));

    cout << ans << endl;
    for(int i=0; i<steps.size(); i++) {
        cout << steps[i].first << " " << steps[i].second << endl;
    }

    
    
    return 0;    
}    

/******************************/    
/***** DECRARED FUNCTIONS *****/    
/******************************/    

int sum_digits(int num) {    
    int rtn = 0;    
    while(num != 0) {    
        int tmp = num % 10;    
        rtn += tmp;    
        num /= 10;    
    }    
    return rtn;    
}    

// vectorのコピー    
template <class T>    
vector<T> copy_vector(vector<T> vec) {    
    vector<T> rtn;    
    rtn.reserve(vec.size());    
    copy(vec.begin(), vec.end(), back_inserter(rtn));    
    return rtn;    
}    

// ユークリウッドの互除法    
int gcd(int a, int b) {    
    if(b==0) return a;    
    else return gcd(b, a % b);    
}