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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include<bits/stdc++.h>
using namespace std;

#define LL long long

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n; cin >> n; 
    int a[n];
    for(int &i : a) cin >> i;
    sort(a, a + n, greater<int>());

    int j = n - 1;
    for(int i = 1; i < n; i++) if(a[i] < 0) {j = i; break;}

    int ans = 0;
    for(int i = 0; i < j; i++) ans += a[i];
    for(int i = j; i < n; i++) ans -= a[i];

    cout << ans << '\n';
    for(int i = 1; i < j; i++) cout << a[j] << " " << a[i] << '\n', a[j] -= a[i];
    for(int i = j; i < n; i++) cout << a[0] << " " << a[i] << '\n', a[0] -= a[i];     
}