description: Theofanis has a riddle for you and if you manage to solve it, he will give you a Cypriot snack halloumi for free (Cypriot cheese).
  
  You are given an integer n. You need to find two integers l and r such that -10^{18} ≤ l < r ≤ 10^{18} and l + (l + 1) + … + (r - 1) + r = n.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The first and only line of each test case contains a single integer n (1 ≤ n ≤ 10^{18}).
  
  Output
  
  For each test case, print the two integers l and r such that -10^{18} ≤ l < r ≤ 10^{18} and l + (l + 1) + … + (r - 1) + r = n. 
  
  It can be proven that an answer always exists. If there are multiple answers, print any.
  
  Example
  
  Input
  
  
  7
  1
  2
  3
  6
  100
  25
  3000000000000
  
  
  Output
  
  
  0 1
  -1 2 
  1 2 
  1 3 
  18 22
  -2 7
  999999999999 1000000000001
  
  Note
  
  In the first test case, 0 + 1 = 1.
  
  In the second test case, (-1) + 0 + 1 + 2 = 2.
  
  In the fourth test case, 1 + 2 + 3 = 6.
  
  In the fifth test case, 18 + 19 + 20 + 21 + 22 = 100.
  
  In the sixth test case, (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 25.
solution: import java.util.*;
  import java.io.*;
  import java.lang.*;
  import java.math.*;
          
           
           
  public class test2{
             public static boolean cp(int a) {
                for(int j=2;j*j<=a;j++)
                {
                   if(a%j==0)
                   {
                      return false;
                   }
                }
                return true;
             }
    public static int gcd(int a, int b) 
      { 
      if (a == 0) 
          return b;  
      return gcd(b % a, a);  
      } 
        
  
            public static void main (String [] args) throws IOException
            {
             
          	  
          	  try {
          		  Scanner sc=new Scanner(System.in);
          		  
          		     int test=sc.nextInt();
          		     
          		     for(int t=0;t<test;t++)
          		     {
          		       long n=sc.nextLong();
          		       
          		       System.out.println((-n+1)+" "+n);
          		     }
  
                    
          	  }
          	  
          	  catch (Exception e) {
          		return;
          	}
           
          }
           
          }  