Author : Kamonasish Roy
Editorial :

• You are given three integers: l, r, and x (with l ≤ r).
• Your task is to find two integers p and q such that:

• l ≤ p ≤ r

• l ≤ q ≤ r

• p - q = x

• If there are multiple valid pairs, any one can be printed. If no valid pair exists, output "-1 -1".

• Key Observations & Approach:

• Case: x = 0

When x is zero, the condition becomes p - q = 0, i.e., p = q.

In this case, any number within [l, r] works.

Solution: Choose p = l and q = l.

• Case: x > 0

We need p - q = x, so p must be greater than q.

A natural choice is to let q be the smallest possible value (q = l).

Then, set p = l + x.

Validity Check: p must not exceed r, i.e., ensure l + x ≤ r.

Solution: If l + x ≤ r, output p = l + x and q = l.

• Case: x < 0

Here, p - q = x (a negative number) implies that p < q.

A convenient choice is to let q be the largest possible value (q = r).

Then, set p = r + x (since x is negative, p will be less than r).

Validity Check: p must be at least l, i.e., ensure r + x ≥ l.

• Solution: If r + x ≥ l, output p = r + x and q = r.

• No Valid Pair:

If neither of the above conditions (for x > 0 or x < 0) are satisfied, then no valid pair (p, q) exists.

Solution: Output "-1 -1".

• Example:

For l = 5, r = 10, x = 3 (x > 0):

Set q = 5 and p = 5 + 3 = 8.

Check: 8 ≤ 10 is true, so output: 8 5.

For l = 5, r = 10, x = -3 (x < 0):

Set q = 10 and p = 10 - 3 = 7.

Check: 7 ≥ 5 is true, so output: 7 10.

Code (Python):

# Author: Kamonasish Roy (Bullet)
# Time: 2025-03-24 04:20:12

def solve():
    t = int(input())  # Number of test cases
    for _ in range(t):
        l, r, x = map(int, input().split())
        
        if abs(x) > (r - l):
            print(-1, -1)
            continue
        
        p = l
        q = l + abs(x)
        
        if x > 0:
            p, q = q, p
        
        print(p, q)

# Call the function to solve the problem
solve()