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Contest: Task: Related: TaskD TaskF

Score : $500$ points

Problem Statement

A train goes back and forth between Town $A$ and Town $B$. It departs Town $A$ at time $0$ and then repeats the following:

  • goes to Town $B$, taking $X$ seconds;
  • stops at Town $B$ for $Y$ seconds;
  • goes to Town $A$, taking $X$ seconds;
  • stops at Town $A$ for $Y$ seconds.

More formally, these intervals of time are half-open, that is, for each $n = 0, 1, 2, \dots$:

  • at time $t$ such that $(2X + 2Y)n ≤ t < (2X + 2Y)n + X$, the train is going to Town $B$;
  • at time $t$ such that $(2X + 2Y)n + X ≤ t < (2X + 2Y)n + X + Y$, the train is stopping at Town $B$;
  • at time $t$ such that $(2X + 2Y)n + X + Y ≤ t < (2X + 2Y)n + 2X + Y$, the train is going to Town $A$;
  • at time $t$ such that $(2X + 2Y)n + 2X + Y ≤ t < (2X + 2Y)(n + 1)$, the train is stopping at Town $A$.

Takahashi is thinking of taking this train to depart Town $A$ at time $0$ and getting off at Town $B$. After the departure, he will repeat the following:

  • be asleep for $P$ seconds;
  • be awake for $Q$ seconds.

Again, these intervals of time are half-open, that is, for each $n = 0, 1, 2, \dots$:

  • at time $t$ such that $(P + Q)n ≤ t < (P + Q)n + P$, Takahashi is asleep;
  • at time $t$ such that $(P + Q)n + P ≤ t < (P + Q)(n + 1)$, Takahashi is awake.

He can get off the train at Town $B$ if it is stopping at Town $B$ and he is awake.
Determine whether he can get off at Town $B$. If he can, find the earliest possible time to do so.
Under the constraints of this problem, it can be proved that the earliest time is an integer.

You are given $T$ cases. Solve each of them.

Constraints

  • All values in input are integers.
  • $1 ≤ T ≤ 10$
  • $1 ≤ X ≤ 10^9$
  • $1 ≤ Y ≤ 500$
  • $1 ≤ P ≤ 10^9$
  • $1 ≤ Q ≤ 500$

Input

Input is given from Standard Input in the following format:

$T$
$\rm case_1$
$\rm case_2$
$\hspace{9pt}\vdots$
$\rm case_T$

Each case is in the following format:

$X$ $Y$ $P$ $Q$

Output

Print $T$ lines.
The $i$-th line should contain the answer to $\rm case_i$.
If there exists a time such that Takahashi can get off the train at Town $B$, the line should contain the earliest such time; otherwise, the line should contain infinity.

Sample Input 1

3
5 2 7 6
1 1 3 1
999999999 1 1000000000 1

Sample Output 1

20
infinity
1000000000999999999

Let $[a, b)$ denote the interval $a ≤ t < b$.

In the first case, the train stops at Town $B$ during $[5, 7), [19, 21), [33, 35), \dots$, and Takahashi is awake during $[7, 13), [20, 26), [33, 39), \dots$, so he can get off at time $20$ at the earliest.