Problem:
Score : 200 points

Problem Statement

There are N pieces of source code. The characteristics of the i-th code is represented by M integers A_{i1}, A_{i2}, ..., A_{iM}.

Additionally, you are given integers B_1, B_2, ..., B_M and C.

The i-th code correctly solves this problem if and only if A_{i1} B_1 + A_{i2} B_2 + ... + A_{iM} B_M + C > 0.

Among the N codes, find the number of codes that correctly solve this problem.

Constraints

All values in input are integers.

1 \leq N, M \leq 20

-100 \leq A_{ij} \leq 100

-100 \leq B_i \leq 100

-100 \leq C \leq 100

Input

Input is given from Standard Input in the following format:

N M C
B_1 B_2 ... B_M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
\vdots
A_{N1} A_{N2} ... A_{NM}

Output

Print the number of codes among the given N codes that correctly solve this problem.

Sample Input 1

2 3 -10
1 2 3
3 2 1
1 2 2

Sample Output 1

1

Only the second code correctly solves this problem, as follows:

Since 3 \times 1 + 2 \times 2 + 1 \times 3 + (-10) = 0 \leq 0, the first code does not solve this problem.

1 \times 1 + 2 \times 2 + 2 \times 3 + (-10) = 1 > 0, the second code solves this problem.

Sample Input 2

5 2 -4
-2 5
100 41
100 40
-3 0
-6 -2
18 -13

Sample Output 2

2

Sample Input 3

3 3 0
100 -100 0
0 100 100
100 100 100
-100 100 100

Sample Output 3

0

All of them are Wrong Answer. Except yours.

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn get_input() -> String {
    let mut word = String::new();
    std::io::stdin().read_line(&mut word).unwrap();
    word.trim().to_owned()
}

fn main() {
    let (n,m,c) = {
        let line = get_input();
        let mut split = line.split_whitespace();
        let n:usize = split.next().unwrap().parse().unwrap();
        let m:usize = split.next().unwrap().parse().unwrap();
        let c:i64 = split.next().unwrap().parse().unwrap();
        (n,m,c)
    };

    let b:Vec<i64> = get_input().split_whitespace().map(|x| x.parse().unwrap()).collect();

    let mut a = vec![vec![0;m]; n];
    for i in 0..n {
        a[i] = get_input().split_whitespace().map(|x| x.parse().unwrap()).collect();
    }

    let mut ans = 0;
    for i in 0..n {
        let s = (0..m).map(|x| a[i][x] * b[x]).collect::<Vec<i64>>().iter().sum::<i64>();
        if s + c > 0 {
            ans += 1;
        }
    }
    println!("{}", ans);
}