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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		//List<String> list= new ArrayList<String>(Arrays.asList(s.split("")));
		List<Integer> list=new ArrayList<Integer>();
		int n = sc.nextInt();
		int m = sc.nextInt();
		int c = sc.nextInt();
		Integer[][] array = new Integer[n][m];
		for(int i=0;m>i;i++) {
			int a = sc.nextInt();
			list.add(a);
		}
		for(int i=0;n>i;i++) {
			for(int j=0;m>j;j++) {
				int a = sc.nextInt();
				array[i][j]=a;
			}
		}
		int count=0;
		for(int i=0;n>i;i++) {
			int keisan=0;
			for(int j=0;m>j;j++) {
				keisan=keisan+list.get(j)*array[i][j];
			}
			if(keisan+c>0)
				count++;
		}
		System.out.println(count);
	}
}