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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <random>
#include <regex>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <sys/timeb.h>
#include <vector>

using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define reprrev(i, a, b) for (int i = (int)(b)-1; i >= (int)(a); i--)
#define reprev(i, n) reprrev(i, 0, n)
#define repi(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)
#define chmin(mi, val) mi = min(mi, val)
#define chmax(ma, val) ma = max(ma, val)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define mp make_pair
#define mt make_tuple
#define INF 1050000000
#define INFR INT_MAX
#define INFL (long long)(4e18)
#define INFLR LLONG_MAX
#define EPS (1e-10)
#define MOD 1000000007
//#define MOD 998244353
#define PI 3.141592653589793238
#define RMAX 4294967295

typedef long long ll;
typedef pair<int, int> Pi;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<Pi> vPi;
typedef vector<Pll> vPll;
typedef vector<Pd> vPd;
typedef vector<vector<int>> vvi;
typedef vector<vector<bool>> vvb;
typedef vector<vector<ll>> vvll;
typedef vector<vector<char>> vvc;
typedef vector<vector<string>> vvs;
typedef vector<vector<double>> vvd;
typedef vector<vector<Pi>> vvPi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<ll>>> vvvll;
typedef vector<vector<vector<Pi>>> vvvPi;
typedef vector<vector<vector<vector<Pi>>>> vvvvPi;
typedef priority_queue<int, vector<int>, greater<int>> pqli;
typedef priority_queue<ll, vector<ll>, greater<ll>> pqlll;
typedef priority_queue<Pi, vector<Pi>, greater<Pi>> pqlP;
template <class T>
using vec = vector<T>;
template <class T>
using pql = priority_queue<T, vector<T>, greater<T>>;

// vvvvvvvvvvvvvvvvvvvvvvv debug output vvvvvvvvvvvvvvvvvvvvvvv
// vector
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec) is >> x;
    return is;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
    os << "(" << pair_var.first << "," << pair_var.second << ")";
    return os;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &pair_var) {
    os << "(" << pair_var.first << "," << pair_var.second << ")";
    return os;
}
// vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}
// map
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
    os << "{";
    repi(itr, map_var) {
        os << "(" << itr->first << "," << itr->second << ")";
        itr++;
        if (itr == map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}
// set
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
    os << "{";
    repi(itr, set_var) {
        os << "  " << *itr << endl;
        itr++;
        if (itr == set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}

#define DUMPOUT cerr

void dump_func() {
    DUMPOUT << endl;
}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define dump(...)                                                              \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                            \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]"        \
            << endl                                                            \
            << "    ",                                                         \
        dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define dump(...)
#endif

// ^^^^^^^^^^^^^^^^^^^^^^^ debug output ^^^^^^^^^^^^^^^^^^^^^^^

string YN(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "YES" : "NO");
}
string yn(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "Yes" : "No");
}
string ON(bool y, int id = 0) {
    if (id) cout << id;
    return (y ? "OK" : "NG");
}

int dir4[4][2] = {{0, -1}, {-1, 0}, {1, 0}, {0, 1}};
int dir8[8][2] = {{-1, -1}, {0, -1}, {1, -1}, {-1, 0},
                  {1, 0},   {-1, 1}, {0, 1},  {1, 1}};
char dirchar[4] = {'<', '^', '>', 'v'};

// [a,b)
int irand(int a, int b) {
    static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
    uniform_int_distribution<int> dist(a, b - 1);
    return dist(Rand);
}

// [a,b)
double drand(int a, int b) {
    static mt19937 Rand(static_cast<unsigned int>(time(nullptr)));
    uniform_real_distribution<double> dist(a, b);
    return dist(Rand);
}

// https://qiita.com/IgnorantCoder/items/3101d6276e9bdddf872c
template <typename A, typename F>
inline auto transform(const A &v, F &&f) {
    using result_type =
        decltype(std::declval<F>()(std::declval<typename A::value_type>()));
    vector<result_type> y(v.size());
    std::transform(std::cbegin(v), std::cend(v), std::begin(y), f);
    return y;
}

// 多次元vector生成
template <class T>
vector<T> make_v(size_t size, const T &init) {
    return vector<T>(size, init);
}
template <class... Ts>
auto make_v(size_t size, Ts... rest) {
    return vector<decltype(make_v(rest...))>(size, make_v(rest...));
}

template <typename T>
T max(vector<T> a) {
    return *max_element(all(a));
}
template <typename T>
T min(vector<T> a) {
    return *min_element(all(a));
}
template <typename T>
T sum(vector<T> a) {
    return accumulate(all(a), (T)0);
}

// mapでカウントとかする
template <typename T>
void add(map<T, int> &m, T item) {
    if (m.find(item) == m.end()) {
        m[item] = 1;
    } else {
        m[item]++;
    }
}

// デフォルト値つきのmapのget
template <typename T, typename U>
U get(map<T, U> m, T key, U def) {
    if (m.find(key) == m.end()) {
        return def;
    } else {
        return m[key];
    }
}

template <typename T>
bool contains(set<T> t, const T &key) {
    return t.find(key) != t.end();
}

struct Edge {
    int from, to, cost;
    bool operator<(Edge e) {
        return cost < e.cost;
    }
};

using Graph = vec<vec<Edge>>;

ostream &operator<<(ostream &os, Edge &edge) {
    os << "(" << edge.from << "->" << edge.to << ":" << edge.cost << ")";
    return os;
}

//======================================================

int main(void) {
    int N, M, C;
    cin >> N >> M >> C;
    vi b(M);
    cin >> b;
    int ans = 0;
    rep(i, N) {
        vi a(M);
        cin >> a;
        int s = C;
        rep(i, M) s += a[i] * b[i];
        if (s > 0) ans++;
    }
    cout << ans << endl;
    return 0;
}