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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define INF -100000

typedef struct data{
  int num;
  int key;
} cell;

int a[100005],v[100005],w[100005];
cell posi[100005],nega[100005];

int main(void){
  int M,i,s,t,max,maxi,min,mini,r;
  char str[1000000],*p;
  
  fgets(str,sizeof(str),stdin);
  M=atoi(str);
  
  fgets(str,sizeof(str),stdin);
  p=strtok(str," \n");
  a[1]=atoi(p);
  for(i=2;i<M+1;i++){
    p=strtok(NULL," \n");
    a[i]=atoi(p);
  }
  
  for(i=1,max=INF;i<M+1;i++){
    if(max<a[i]){
      max=a[i],maxi=i;
    }
  }
  for(i=1,min=100000;i<M+1;i++){
    if(min>a[i]){
      min=a[i],mini=i;
    }
  }
  
  for(i=1,s=0,t=0;i<M+1;i++){
    if(i!=maxi){
      if(a[i]>=0){
      posi[s].num=a[i],posi[s].key=i,s++;
      }else{
        nega[t].num=a[i],nega[t].key=i,t++;
      }
    }
  }
  r=0;
  
  if(t==0){
    while(s>0){
      s--;
      if(s!=mini){
        v[r]=min,w[r]=posi[s].num,r++;
        min-=posi[s].num;
      }
    }
    v[r]=max,w[r]=min,r++;
    max-=min;
    
    printf("%d\n",max);
    for(i=0;i<r;i++){
      printf("%d %d\n",v[i],w[i]);
    }
    
    return 0;
  }
  
  if(s==0){
      while(t>0){
        t--;
        v[r]=max,w[r]=nega[t].num,r++;
        max-=nega[t].num;
  	  }
    printf("%d\n",max);
     for(i=0;i<r;i++){
      printf("%d %d\n",v[i],w[i]);
    }
    return 0;
  }
  
  while(s>0){
    s--;
    v[r]=nega[0].num,w[r]=posi[s].num,r++;
    nega[0].num-=posi[s].num;
  }
  
  while(t>0){
    t--;
    v[r]=max,w[r]=nega[t].num,r++;
    max-=nega[t].num;
    
  }
  printf("%d\n",max);
  
  for(i=0;i<r;i++){
    printf("%d %d\n",v[i],w[i]);
  }
  return 0;
  
}