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.*;
import java.math.*;
import java.util.Scanner;
import java.math.BigInteger;

public class Main {
	static final long mod = (int) (1e9 + 7);
	static final int inf = (int) (1e9);
	static final int maxn = (int) (1e5 + 10);

	static int n, m,c;
	static String s;
	static int[][] a=new int [22][22];
	static int [] b=new int[22];
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		n=cin.nextInt();m=cin.nextInt();c=cin.nextInt();
		for(int i=1;i<=m;i++) {
			b[i]=cin.nextInt();
		}
		int sum=0,ans=0;
		for(int i=1;i<=n;i++) {
			sum=c;
			for(int j=1;j<=m;j++) {
				a[i][j]=cin.nextInt();
				sum+=a[i][j]*b[j];
			}
			if(sum>0) ans++;
		}
		System.out.println(ans);
	}
}