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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;
#include <vector>

int innerProduct(vector<int> a, vector<int> b, int C){
	int ans=0;
	int s = a.size();
	for(int i=0; i<s; i++){
		ans += a[i]*b[i];
		
	}
	ans += C;
	return ans;
}

int main() {
	int N, M, C;
	
	cin >> N >> M >> C;
	
	//2次元配列
	vector<vector<int>> v(N, vector<int>(M));
	vector<int> B(M);
	
	for(int i=0; i<M; i++){
		cin >> B[i];
	}
	
	//N回M個のデータを受け取る
	for(int i=0; i< N; i++){
		for(int j=0; j<M; j++){
			cin >> v[i][j];
		}
	}
	
	int cnt=0;
	for(int i=0; i<N; i++){
		int ans = innerProduct(B, v[i], C);
		cout << "ans is... " << ans;
		cout << endl;
		if(ans > 0) cnt++;
	}
	
	cout << cnt;
	cout << endl;
	return 0;
}