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<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
#define REP(i, limit) for(int i=0; i< limit; ++i)
#define FOR(i, j, limit) for(int i=j; i< limit; ++i)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ECHO(x) cout<<x<<endl;
#define LL long long
int INF = 2147483647;
LL LINF = 9223372036854775807;
LL MOD = 1000000007;
typedef pair<int, int>P;
typedef pair<LL, LL>PL;
typedef vector<int> vi;
typedef vector<LL> vl;


int main(){
  int n, m, c;
  int ans=0;
  cin>>n>>m>>c;
  int b[m], a[n][m];
  REP(i, m)cin>>b[i];
  REP(i, n)REP(j, m)cin>>a[i][j];

  REP(i, n){
    int work=0;
    REP(j, m) work+= a[i][j]*b[j];
    work+=c;
    if(work>0) ++ans;
  }
  cout<<ans<<endl;

  return 0;
}