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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		String[] line = sc.nextLine().split(" ");
		int N = Integer.parseInt(line[0]);
		int M = Integer.parseInt(line[1]);
		int C = Integer.parseInt(line[2]);
		
		line = sc.nextLine().split(" ");
		int[] B = new int[M];
		
		for(int i= 0;i<M;i++) {
			B[i] = Integer.parseInt(line[i]);
		}
		
		int[][] A = new int[N][M];
		for(int i= 0;i<N;i++) {
			line = sc.nextLine().split(" ");
			for(int j= 0;j<M;j++) {
				A[i][j] = Integer.parseInt(line[j]);
			}
		}
		sc.close();
		
		int count =0;
		
		for(int i= 0;i<N;i++) {
			int sum = C;
			for(int j= 0;j<M;j++) {
				sum += A[i][j] * B[j];
			}
			if (sum>0) {
				count ++;
			}
		}
		
		System.out.println(count);

	}

}