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[]){
		  
	    Scanner sc = new Scanner(System.in);
	    
	    int N = sc.nextInt();
	    int M = sc.nextInt();
	    int C = sc.nextInt();
	    
	    int B[] = new int[M];
	    int A[][] = new int[N][M];
	    
	    for(int i = 0; i < B.length; i++) {
	    	B[i] = sc.nextInt();
	    }
	    
	    for(int i = 0; i < N; i++) {
	    	for (int j = 0; j < M; j++) {
				A[i] [j] = sc.nextInt();
			}
	    }
	    
	    int cnt = 0;
	    int sum = 0;
	    for(int i = 0; i < N; i++) {
	    	for(int j = 0; j < B.length; j++) {
	    		sum += A[i][j]*B[j];
	    	}
	    	if(sum+C > 0) {
	    		cnt++;
	    	}
	    	sum = 0;
	    }
	    
	    System.out.println(cnt);
	    
	    
	}
	
}