Consecutive Sum Riddle codeforces solution- 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

Consecutive Sum Riddle codeforces solution

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 1018l<r1018−1018≤l<r≤1018 and l+(l+1)++(r1)+r=nl+(l+1)+…+(r−1)+r=n.

Consecutive Sum Riddle solution codeforces

The first line contains a single integer tt (1t1041≤t≤104) — the number of test cases.

The first and only line of each test case contains a single integer nn (1n10181≤n≤1018).

Consecutive Sum Riddle solution codeforces

For each test case, print the two integers ll and rr such that 1018l<r1018−1018≤l<r≤1018 and l+(l+1)++(r1)+r=nl+(l+1)+…+(r−1)+r=n.

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

Example

input

Consecutive Sum Riddle solution codeforces

7
1
2
3
6
100
25
3000000000000

Consecutive Sum Riddle solution codeforces

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

Consecutive Sum Riddle solution codeforces

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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