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 <algorithm>
#include <vector>
#include <string.h>
#include <cmath>
#include <queue>
using namespace std;
#define mod (long)(1e9 + 7)
#define all(x) (x).begin(), (x).end()
#define bitcount(n) __builtin_popcountl(long(n))
#define fcout cout << fixed << setprecision(15)
#define highest(x) (63 - __builtin_clzl(x))
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void POSS(T condition){ if(condition) cout << "POSSIBLE" << endl; else cout << "IMPOSSIBLE" << endl; }
template<class T> inline void Poss(T condition){ if(condition) cout << "Possible" << endl; else cout << "Impossible" << endl; }
template<class T> inline void First(T condition){ if(condition) cout << "First" << endl; else cout << "Second" << endl; }
template<class T = string, class U = char>int character_count(T text, U character){ int ans = 0; for(U i: text){ ans += (i == character); } return ans; }
long power(long base, long exponent, long module){ if(exponent % 2){ return power(base, exponent - 1, module) * base % module; }else if(exponent){ long root_ans = power(base, exponent / 2, module); return root_ans * root_ans % module; }else{ return 1; }}
struct position{ int y, x; }; position mv[4] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // double euclidean(position first, position second){ return sqrt((second.x - first.x) * (second.x - first.x) + (second.y - first.y) * (second.y - first.y)); }
template<class T, class U> string to_string(pair<T, U> x){ return to_string(x.first) + "," + to_string(x.second); } string to_string(string x){ return x; }
template<class itr> void array_output(itr start, itr goal){ string ans; for(auto i = start; i != goal; i++) ans += to_string(*i) + " "; if(!ans.empty()) ans.pop_back(); cout << ans << endl; }
template<class itr> void cins(itr start, itr goal){ for(auto i = start; i != goal; i++){ cin >> (*i); } }
template<class T> T gcd(T a, T b){ if(a && b){ return gcd(min(a, b), max(a, b) % min(a, b)); }else{ return a; }} template<class T> T lcm(T a, T b){ return a / gcd(a, b) * b; }
struct combination{ vector<long> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2, mod); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } long C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

int main(){
    int N, M, C;
    cin >> N >> M >> C;
    int B[M];
    cins(B, B + M);
    int ans = 0;
    for(int i = 0; i < N; i++){
        int A[M];
        cins(A, A + M);
        int sum = C;
        for(int j = 0; j < M; j++){
            sum += A[j] * B[j];
        }
        if(sum > 0){
            ans++;
        }
    }
    cout << ans << endl;
}