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 <stdio.h>
#include <string>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <limits.h>
#include <map>
#include <numeric>
#include <queue>

namespace test {
  std::string to_string(char val) {
    return std::string(1, val);
    // return std::string{val};
  }
}

using namespace std;



int main(void) {
  long long int n,m,c; cin >> n >> m >>c;
  long long int *b = new long long int[m];
  long long int *a = new long long int[m];
  int result = 0;
  long long int total = 0;
  for(int i = 0; i< n+1; i++){
    for(int j =0; j < m; j++){
      if(i == 0){
        cin >> b[j];
      } else {
        cin >> a[j];
        total += a[j] * b[j];
      }
    }
    total += c;
    if(i != 0 && total > 0){
      result++;
    }
    
    total = 0;
  }

  cout << result << endl;

  return 0;
}