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 <bits/stdc++.h>
using namespace std;

using lli = long long int;
using Vint = vector<int>;
using Vlli = vector<lli>;
using Wint = vector<Vint>;
using Wlli = vector<Vlli>;
using pii = pair<int, int>;
using pll = pair<lli, lli>;

const int MOD = 1e9 + 7;
const int INFi = 2e9 + 1;
const lli INFl = (lli)(9e18) + 1;
const vector<pii> DXDY = {make_pair(1, 0), make_pair(-1, 0), make_pair(0, 1), make_pair(0, -1)};

#define FOR(i, a, b) for(int (i) = (a); (i) < (b); (i)++)
#define FOReq(i, a, b) for(int (i) = (a); (i) <= (b); (i)++)
#define rFOR(i, a, b) for(int (i) = (b); (i) >= (a); i--)
#define FORstep(i, a, b, step) for(int (i) = (a); i < (b); i += (step))
#define REP(i, n) FOR(i, 0, n)
#define rREP(i, n) rFOR(i, 0, (n-1))
#define vREP(i, vec) for(auto &(i) : (vec))


template <class T> inline int argmin(vector<T> vec){return min_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline int argmax(vector<T> vec){return max_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline void chmax(T &a, T b){a = max(a, b);}
template <class T> inline void chmin(T &a, T b){a = min(a, b);}
inline int toInt(string &s){int res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline int toInt(const string s){int res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline long long int toLong(string &s){lli res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline long long int toLong(const string s){lli res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
template <class T> inline std::string toString(T n){
  if(n == 0) return "0";
  std::string res = "";
  if(n < 0){n = -n;while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return '-' + res;}
  while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return res;
}

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


int main(void){
  int n, m, c, res = 0;
  cin >> n >> m >> c;
  Vint B(m);
  vREP(ele, B) cin >> ele;
  int SUM;
  REP(i, n){
    SUM = 0;
    int tmp;
    REP(j, m){
      cin >> tmp;
      SUM += B.at(j) * tmp;
    }
    if(SUM + c > 0) res++;
  }
  cout << res << '\n';
  return 0;
}