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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    // H,W,h,wを取得
    int N = sc.nextInt(); //列数
    int M = sc.nextInt(); //行数
    int C = sc.nextInt(); //const
    int[][] A = new int[N][M];
    int[][] B = new int[M][1];
    for(int i=0;i<M;i++){
      B[i][0] = sc.nextInt();
    }
    for(int i=0;i<N;i++){
      for(int j=0;j<M;j++){
        A[i][j] = sc.nextInt();
      }
    }
    // 行列計算
    int[][] ansMat = new int[M][1];
    ansMat = multArray(A,B);
    // -c以上の数だけインクリメント
    int count = 0;
    // System.out.println(ansMat.length+" "+ansMat[0].length);
    for(int i=0;i<N;i++){
      if(ansMat[i][0]>-C){
        count++;
      }
    }
    // 出力
    System.out.println(count);
  }
  public static int[][] multArray(int a[][], int b[][]){
    int ay = a.length;
    int ax = a[0].length;
    int by = b.length;
    int bx = b[0].length;
    int[][] c = new int[ay][bx];
    if(ax!=by){
      return c;
    }
    // 行列積
    for(int i=0;i<ay;i++){
      for(int j=0;j<bx;j++){
        int sum = 0;
        for(int k=0;k<ax;k++){
          sum += a[i][k]*b[k][j];
        }
        c[i][j] = sum;
      }
    }
    return c;
  }
}