[Solution] Consecutive Sum Riddle solution codeforces

Consecutive Sum Riddle solution codeforces


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 𝑛n. You need to find two integers 𝑙l and 𝑟r such that −1018≤𝑙<𝑟≤1018−1018≤l<r≤1018 and 𝑙+(𝑙+1)+…+(𝑟−1)+𝑟=𝑛l+(l+1)+…+(r−1)+r=n.

Consecutive Sum Riddle solution codeforces

Input

The first line contains a single integer 𝑡t (1≤𝑡≤1041≤t≤104) — the number of test cases.

The first and only line of each test case contains a single integer 𝑛n (1≤𝑛≤10181≤n≤1018).

Consecutive Sum Riddle solution codeforces

Output

For each test case, print the two integers 𝑙l and 𝑟r such that −1018≤𝑙<𝑟≤1018−1018≤l<r≤1018 and 𝑙+(𝑙+1)+…+(𝑟−1)+𝑟=𝑛l+(l+1)+…+(r−1)+r=n.

Consecutive Sum Riddle solution codeforces

It can be proven that an answer always exists. If there are multiple answers, print any.

Example
input

Copy
7
1
2
3
6
100
25
3000000000000

Consecutive Sum Riddle solution codeforces

 

output

Copy
0 1
-1 2 
1 2 
1 3 
18 22
-2 7
999999999999 1000000000001

Consecutive Sum Riddle solution codeforces

Note

In the first test case, 0+1=10+1=1.

In the second test case, (−1)+0+1+2=2(−1)+0+1+2=2.

In the fourth test case, 1+2+3=61+2+3=6.

In the fifth test case, 18+19+20+21+22=10018+19+20+21+22=100.

In the sixth test case, (−2)+(−1)+0+1+2+3+4+5+6+7=25(−2)+(−1)+0+1+2+3+4+5+6+7=25.


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