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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repb(i, a, b) for (int i = (a); i < (b); i++)
#define ALL(a) (a).begin(),(a).end()
#define PRINT(V) for(auto v:(V))cout<<v<<" "<<endl
inline int GCD(int a, int b){return b ? GCD(b, a % b) : a;}
const int MOD = 1000000007;

int main()
{
    int N, M, C;
    cin >> N >> M >> C;
    vi B(M);
    rep(i, M)
    {
        cin >> B.at(i);
    }
    vvi A(N, vi(M));
    rep(i, N)
    {
        rep(j, M)
        {
            cin >> A.at(i).at(j);
        }
    }
    int ans = 0;
    rep(i, N)
    {
        int sum = 0;
        rep(j, M)
        {
            sum += A.at(i).at(j) * B.at(j);
        }
        sum += C;
        if (sum > 0)
        {
            ans++;
        }
    }
    cout << ans << endl;
}