Table of Contents

## 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

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

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.

7 1 2 3 6 100 25 3000000000000

### Consecutive Sum Riddle solution codeforces

output

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.