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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

int main()
{
 
  int a, b, c, d, e, X, Y;
  //  double  c;
  int  A[101][101]={};
  int  B[100001]={};
//  int  B[100003]={};
  int  L[1003]={};

  int  flag[103]={};

//  long long  a, b, c, d, X, Y;
char s[100001]={};
  long long JPY;
  double BTC;
char t[100001]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
//  int flag=0;
  double tmp=0;
//  int min=1000000;
  long long min=100000001;
  int len=0;
  int cnt=0;
  int cnt0=0;
  int cnt1=0;
  int num=0;
//  double ans=0.0;
  long long ans[100001]={};
  int max=0;
  int sum=0;
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%d%d%d", &a,&b, &c);
  for(i=0;i<a+1;i++){
    for(j=0;j<b;j++){
      if(i==0)  scanf("%d", &B[j]);
      else  scanf("%d", &A[i-1][j]);
    }
  }
  for(i=0;i<a;i++){
    for(j=0;j<b;j++){
      sum=sum+A[i][j]*B[j];
    }
    if(sum>-c) cnt++;
    sum=0;
  }      
  
 printf("%d\n", cnt);


  //    if(max < cnt - max)  printf("Yes\n");
//    else  
  
return 0;
}