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 JavaScript (node.js v5.12) (judge verdict: Accepted)

function main(input) {
  const formatted_input = input.split('\n');
  const N =  parseInt(formatted_input[0].split(' ')[0]);
  const M =  parseInt(formatted_input[0].split(' ')[1]);
  const C =  parseInt(formatted_input[0].split(' ')[2]);
  const Bs = formatted_input[1].split(' ').map(x => parseInt(x));

  As=[];
  for(var i=0;i<N;i++){
    var tmp_as = formatted_input[2+i].split(' ').map(x => parseInt(x))
    As.push(tmp_as);
  }

//  console.log(N);
//  console.log(M);
//  console.log(C);
//  console.log(Bs);
//  console.log(As);


  var result = 0;

  for(var i=0;i<N;i++){
    var tmp_sum = C;
    for(var j=0;j<M;j++){
      tmp_sum += Bs[j] * As[i][j];
    }

    if(tmp_sum>0){result++;}
  }

  console.log(result);

}

main(require('fs').readFileSync('/dev/stdin', 'utf8'));