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)

use std::io::Read;

fn main() {
    let mut buf = String::new();


    std::io::stdin().read_to_string(&mut buf).unwrap();


    let mut iter = buf.split_whitespace();

    let N: i32 = iter.next().unwrap().parse().unwrap();
    let M: i32 = iter.next().unwrap().parse().unwrap();
    let C: i32 = iter.next().unwrap().parse().unwrap();

    let B: Vec<i32> = (0..M).map(|_| { iter.next().unwrap().parse().unwrap() }).collect();
    let A: Vec<Vec<i32>> = (0..N).map(|_| {
        (0..M).map(|_| { iter.next().unwrap().parse().unwrap() }).collect()
    }).collect();
    let mut result = 0;
    for i in 0..N {
        let mut sum = 0;
        for j in 0..M {
            sum += A[i as usize][j as usize] * B[j as usize];
        }
        sum += C;
        if sum > 0 {
            result += 1;
        }
    }
    println!("{}", result);
}