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 <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime> 
#include <memory.h>
#include <assert.h>

using namespace std;

using ll = long long;
using pii = pair<int, int>;
const int INF = 1e9 + 1;
const ll LLINF = 1e18 + 1;

int main() {
    ll N, M, C;
    cin >> N >> M >> C;
    vector<ll> B(M);
    for(int i = 0; i < M; i++) cin >> B[i];
    vector<vector<ll> > A(N, vector<ll>(M));
    for(int i = 0; i < N; i++) for(int j = 0; j < M; j++) cin >> A[i][j];

    ll ans = 0;
    for(int i = 0; i < N; i++) {
        ll s = 0;
        for(int j = 0; j < M; j++) {
            s += B[j] * A[i][j];
        }
        s += C;
        if(s > 0) ans++;
    }

    cout << ans << endl;

    return 0;
}