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<string>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
using namespace std;
///////////////////library zone!!!!!!!!!!!!!!!!!!!!!!!!!!!!
typedef long long ll;
typedef long double ld;
#define all(a) (a).begin(),(a).end()
const ll Mod = 1000000007;
const ll mod = 0;
struct H {
	ll x, y;
	bool operator<(const H &b) const {
		if (x != b.x) return x < b.x;
		return y < b.y;
	}
	bool operator>(const H &b) const {
		if (x != b.x) return x > b.x;
		return y > b.y;
	}
	bool operator==(const H &b) const { return x == b.x&&y == b.y; }
	bool operator!=(const H &b) const { return (*this) != b; }
};
struct P {
	ll pos, cost;
	bool operator<(const P &b) const { return cost < b.cost; }
	bool operator>(const P &b) const { return cost > b.cost; }
};
struct B {
	ll to, cost;
};
struct E {
	ll from, to, cost;
	bool operator<(const E &b) const { return cost < b.cost; }
	bool operator>(const E &b) const { return cost > b.cost; }
};
template<typename T, typename U>
void chmin(T &a, U b) {
	if (a > b) a = b;
}
template<typename T, typename U>
void chmax(T &a, U b) {
	if (a < b) a = b;
}
template<typename T>
T max_0(T a) {
	if (a < 0) return 0;
	return a;
}
template<typename T>
T min_0(T a) {
	if (a > 0) return 0;
	return a;
}
ll read() {
	ll u;
	scanf("%lld", &u);
	return u;
}
ll gcd(ll i, ll j) {
	if (i > j) swap(i, j);
	if (i == 0) return j;
	return gcd(j%i, i);
}
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}//x^n%p
const ll Inf = 3023372036854775807;
const int inf = 1500000000;
#define int long long
//----------------------------------------------------
int a[2000];
int b[2000];
signed main() {
	int n = read(), m = read(), c = read();
	for (int i = 0; i < m; i++) {
		cin >> b[i];
	}
	int ans = 0;
	for (int i = 0; i < n; i++) {
		int sum = 0;
		for (int j = 0; j < m; j++) {
			sum += read()*b[j];
		}
		if (sum + c > 0) {
			ans++;
		}
	}
	cout << ans << endl;
}