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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <set>
#include <math.h>
#include <cmath>
#define ll long long

using namespace std;

int main(){

  int N, M, C, a, b;
  cin >> N >> M >> C;
  int ans = 0;
  vector<int> arrB(M);
  vector<vector<int>> arrA(N, vector<int>(M));

  for (int i = 0; i < M; i++){
    cin >> b;
    arrB.at(i) = b;
  }

  for (int i = 0; i < N; i++){
    for (int j = 0; j < M; j++) {
      cin >> a;
      arrA.at(i).at(j) = a;
    }
  }

  for (int i = 0; i < N; i++){
    int num = 0;
    for (int j = 0; j < M; j++) {
      num += arrB.at(j) * arrA.at(i).at(j);
    }
    num += C;
    if(num > 0) ans++;
  }

  cout << ans << endl;

  return 0;
}