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;
// type alias
typedef long long LL;
typedef pair<int,int> II;
typedef tuple<int,int,int> III;
typedef vector<int> VI;
typedef vector<string> VS;
typedef unordered_map<int,int> MAPII;
typedef unordered_set<int> SETI;
template<class T> using VV=vector<vector<T>>;
// minmax
template<class T> inline T SMIN(T& a, const T b) { return a=(a>b)?b:a; }
template<class T> inline T SMAX(T& a, const T b) { return a=(a<b)?b:a; }
// repetition
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)
#define REPE(i,n)  for(int i=0;i<=(n);++i)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  for(int i=0;i<(n);++i)
#define FORR(x,arr) for(auto& x:arr)
#define SZ(a) int((a).size())
// collection
#define ALL(c) (c).begin(),(c).end()
// DP
#define MINUS(dp) memset(dp, -1, sizeof(dp))
#define ZERO(dp) memset(dp, 0, sizeof(dp))
// stdout
#define println(args...) fprintf(stdout, ##args),putchar('\n');
// debug cerr
template<class Iter> void __kumaerrc(Iter begin, Iter end) { for(; begin!=end; ++begin) { cerr<<*begin<<','; } cerr<<endl; }
void __kumaerr(istream_iterator<string> it) { (void)it; cerr<<endl; }
template<typename T, typename... Args> void __kumaerr(istream_iterator<string> it, T a, Args... args) { cerr<<*it<<"="<<a<<", ",__kumaerr(++it, args...); }
template<typename S, typename T> std::ostream& operator<<(std::ostream& _os, const std::pair<S,T>& _p) { return _os<<"{"<<_p.first<<','<<_p.second<<"}"; }
#define __KUMATRACE__ true
#ifdef __KUMATRACE__
#define dump(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); __kumaerr(_it, args); }
#define dumpc(ar) { cerr<<#ar<<": "; FORR(x,(ar)) { cerr << x << ','; } cerr << endl; }
#define dumpC(beg,end) { cerr<<"~"<<#end<<": "; __kumaerrc(beg,end); }
#else
#define dump(args...)
#define dumpc(ar)
#define dumpC(beg,end)
#endif

// $ cp-batch CanYouSolveThis | diff CanYouSolveThis.out -
// $ g++ -std=c++14 -Wall -O2 -D_GLIBCXX_DEBUG -fsanitize=address CanYouSolveThis.cpp && ./a.out

/*
 
 3/16/2019
 
 11:14-
 
 */

const int MAX_N=1e2+1;
LL A[MAX_N][MAX_N],B[MAX_N];
int N,M,C;

void solve() {
  int res=0;
  REP(i,N) {
    LL x=C;
    REP(j,M) x+=A[i][j]*B[j];
    if(x>0) ++res;
  }
  cout<<res<<endl;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout<<setprecision(12)<<fixed;
  
  cin>>N>>M>>C;
  REP(i,M) cin>>B[i];
  REP(i,N)REP(j,M) cin>>A[i][j];
  solve();
  
  return 0;
}