Consecutive Sum Riddle
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 $-10^{18} \le l < r \le 10^{18}$ and $l + (l + 1) + \ldots + (r - 1) + r = n$.
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first and only line of each test case contains a single integer $n$ ($1 \le n \le 10^{18}$).
For each test case, print the two integers $l$ and $r$ such that $-10^{18} \le l < r \le 10^{18}$ and $l + (l + 1) + \ldots + (r - 1) + r = n$.
It can be proven that an answer always exists. If there are multiple answers, print any.
Input
The first line contains a single integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first and only line of each test case contains a single integer $n$ ($1 \le n \le 10^{18}$).
Output
For each test case, print the two integers $l$ and $r$ such that $-10^{18} \le l < r \le 10^{18}$ and $l + (l + 1) + \ldots + (r - 1) + r = n$.
It can be proven that an answer always exists. If there are multiple answers, print any.
Samples
7
1
2
3
6
100
25
3000000000000
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, $(-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 25$.