# PROBLEM LINK:

**Setter:** jeevanjyot

**Testers:** tejas10p, rivalq

**Editorialist:** hrishik85

# DIFFICULTY:

1187

# PREREQUISITES:

None

# PROBLEM:

Chef is very hungry. So, Chef goes to a shop selling burgers. The shop has 2 types of burgers:

- Normal burgers, which cost X rupees each
- Premium burgers, which cost Y rupees each (where Y \gt X)

Chef has R rupees. Chef wants to buy **exactly** N burgers. He also wants to maximize the number of premium burgers he buys. Determine the number of burgers of both types Chef must buy.

Output -1 if it is not possible for Chef to buy N burgers.

# EXPLANATION:

The least amount that the Chef definitely needs is (X \times N). If R is less than this amount, we output -1

If the count of burgers purchased are x (normal burgers) and y (premium burgers), then we have the equation R \geq x \times X + y \times Y

R \geq (N - y) \times X + y \times Y

y \leq (R -(N\times X)) \div (Y-X)

Also - y \leq N

Once y is computed, we can output (N-y, y)

# TIME COMPLEXITY:

Time complexity is O(1).

# SOLUTION:

## Editorialist's Solution

```
t=int(input())
for _ in range(t):
X,Y,N,R = map(int,input().split())
if (N*X)>R:
print(-1)
else:
#R=X*(N-y)+Y*y
y = min((R-N*X)//(Y-X),N)
x = N - y
print(x,y)
```