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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;

class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int[] in  = Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    int n = in[0];
    int m = in[1];
    int c = in[2];
    int[] b = Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    int r = 0;
    for (int i = 0; i < n; i++) {
      int[] a = Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
      int sum = 0;
      for (int j = 0; j < m; j++) {
        sum += a[j] * b[j];
      }
      if (sum + c > 0) {
        ++r;
      }
    }
    System.out.println(r);
    sc.close();
  }
}