Generalized Hundred-Chickens Problem

Problem Description

The "Hundred Chickens" problem is a famous mathematical puzzle from the ancient Chinese book Zhang Qiujian Suanjing. In brief:

"Each rooster costs $5$ yuan, each hen costs $3$ yuan, and $3$ chicks cost $1$ yuan; now with $100$ yuan you buy $100$ chickens — how many schemes are there?"

Xiao Ming loves this story and decides to generalize it and solve it by programming: if each rooster costs $x$ yuan, each hen costs $y$ yuan, and every $z$ chicks cost $1$ yuan; and now with $n$ yuan you buy $m$ chickens, how many schemes are there?

Input Format

The input gives one line containing five integers: $x$, $y$, $z$, $n$, $m$ from the problem description. It is given that $1 \le x, y, z \le 10$ and $1 \le n, m \le 1000$.

Output Format

Output one line containing a single integer $C$, the number of schemes.

Sample Input 1

5 3 3 100 100

Sample Output 1

4

Sample Input 2

1 1 1 100 100

Sample Output 2

5151

Sample Explanation

The first sample corresponds to the classic "Hundred Chickens" problem: each rooster $5$ yuan, each hen $3$ yuan, every $3$ chicks $1$ yuan, $100$ yuan total, buying $100$ chickens. There are $4$ schemes:

• $0$ roosters, $25$ hens, $75$ chicks: cost $0 \times 5 + 25 \times 3 + 75 \div 3 = 0 + 75 + 25 = 100$ yuan, total $0 + 25 + 75 = 100$ chickens.
• $4$ roosters, $18$ hens, $78$ chicks: cost $4 \times 5 + 18 \times 3 + 78 \div 3 = 20 + 54 + 26 = 100$ yuan, total $4 + 18 + 78 = 100$ chickens.
• $8$ roosters, $11$ hens, $81$ chicks: cost $8 \times 5 + 11 \times 3 + 81 \div 3 = 40 + 33 + 27 = 100$ yuan, total $8 + 11 + 81 = 100$ chickens.
• $12$ roosters, $4$ hens, $84$ chicks: cost $12 \times 5 + 4 \times 3 + 84 \div 3 = 60 + 12 + 28 = 100$ yuan, total $12 + 4 + 84 = 100$ chickens.

Therefore the output is $4$.

Constraints

$1 \le x, y, z \le 10$, $1 \le n, m \le 1000$.