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)

#include <cstdio>
#include <cmath>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <bitset>
#include <deque>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define print(x) cout<<(x)<<endl
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define INF10 1001001001 
#define INF5 100000
#define EPS 1e-10


int main() {
	int N, M, C; cin >> N >> M >> C;
	vector<int>B(M);
	REP(i, M)cin >> B[i];
	vector<int>A(M);
	int cnt = 0;
	REP(i, N) {
		REP(i, M)cin >> A[i];
		int ans = C;
		REP(i, M)ans += A[i] * B[i];
		if (ans > 0)cnt++;
	}
	print(cnt);

	return 0;
}