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)

/* 
chirag11032000
Chirag Thakur
*/

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

using ll = long long;
using db = double;
using pii = pair < int, int >;
using pll = pair < ll, ll >;

template < typename T > using OrderedSet = tree < T, null_type, less < T >, rb_tree_tag, tree_order_statistics_node_update >;
template < typename T > using MinPriorityQueue = priority_queue < T, vector < T >, greater < T > >;

#ifndef ONLINE_JUDGE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template < typename Arg1 >
void __f(const char* name, Arg1&& arg1) {
	cerr << name << " : " << arg1 << std::endl;
}
template < typename Arg1, typename... Args >
void __f(const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma = strchr(names + 1, ',');
	cerr.write(names, comma - names) << " : " << arg1 << " | ";
	__f(comma + 1, args...);
}
#else
#define trace(...)
#endif

#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define unique_sort(x) sort(all(x)), x.resize(distance(x.begin(), unique(all(x))))

const db PI = acos(-1);
const ll LINF = LLONG_MAX;
const int INF = INT_MAX, MOD = 1e9 + 7, N = 1e5 + 10;

void test_case();

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	test_case();
	cerr << "Time taken: " << int((clock() * 1000.) / CLOCKS_PER_SEC) << "ms\n";
	return 0;
}

void test_case() {
	int n, m, c;
	cin >> n >> m >> c;
	vector < int > b(m);
	for (auto &el: b) {
		cin >> el;
	}
	int cnt = 0;
	while (n--) {
		int tot = 0;
		for (int i = 0; i < m; ++i) {
			int x;
			cin >> x;
			tot += b[i] * x;
		}
		if (tot + c > 0) {
			++cnt;
		}
	}
	cout << cnt << "\n";
}