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: Accepted)

#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <map>
#define rep(i, n) for(int i=0;i<(int)(n);i++)
#define pb(a) push_back(a)
#define MAX(a, b) ((a < b)?b:a)
#define MIN(a, b) ((a > b)?b:a)
typedef long long int ll;
typedef unsigned long long int ull;
using namespace std;

template<typename T> void cinpb(vector<T>& x) { T n; cin >> n; x.push_back(n); }
ull gcd(ull a, ull b) { return b ? gcd(b, a%b) : a; }
ull lcm(ull m, ull n) {	return ((0 == m) || (0 == n)) ? 0 : ((m / gcd(m, n)) * n); }
int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

void yesno(bool f) { cout << (f ? "Yes" : "No") << endl; }

//

int main()
{
	int N, M, C;

	cin >> N >> M >> C;
	vector<int>B;
	int count = 0;

	rep(i, M)
	{
		cinpb(B);
	}

	rep(n, N)
	{
		int sum = 0;
		rep(i, M)
		{
			int A;
			cin >> A;
			sum += A * B[i];
		}
		if ( sum + C > 0 ) count++;
	}

	cout << count << endl;

	return 0;
}