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)

/*

    author : s@if

*/

#include<bits/stdc++.h>
using namespace std;
#define NIL -1
#define fi first
#define sec second
#define MAX INT_MAX
#define INF 99999999
#define ll long long
#define PI acos(-1.0)
#define MOD 1000000007
#define PLL pair<ll,ll>
#define PII pair<int,int>
#define triplell pair<pair<ll,ll> , ll>
#define triple pair<pair<int,int> , int>
#define For(i,n) for(int i=0;i<(int)n;i++)
#define Forn(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define forba(i,a,b) for(int i=(int)b;i>=(int)a;i--)

int dx[]={+1,-1,+0,+0};
int dy[]={+0,+0,+1,-1};
int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};
int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};
int hr[]={-2,-2,-1,+1,+2,+2,-1,+1};
int hc[]={+1,-1,+2,+2,-1,+1,-2,-2};

const int MAXN = (int)1e5+9;

const int N = 100000;

int main()
{
/*	freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);  */
    
    int i,j,k,l,m,n,c,p,q,r,tc,t;
    
    cin>>n>>m>>c;
    
    int ara[m+9];
    for(i=1;i<=m;i++)
    	cin>>ara[i];

	int cnt = 0;
	
	for(i=1;i<=n;i++)
	{
		int s = 0;
		
		for(j=1;j<=m;j++)
		{
			cin>>p;
			
			s+=p*ara[j];
		}
		s+=c;
		
		if(s>0)
			cnt++;
	}
	
	cout<<cnt<<endl;
	
//	main();
    return 0;
}