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;

typedef long long ll;
typedef pair<ll, ll> P;


#define rep(i,n) for(ll (i)=0; (i)<(ll)(n); (i)++)
#define frep(i,m,n) for(ll (i)=(m); (i)<=(ll)(n); (i)++)
#define rrep(i,n) for(ll (i)=(n)-1; (i)>-1; (i)--)
#define frrep(i,m,n) for(ll (i)=(n); (i)>(ll)(m); (i)--)
#define ALL(x) (x).begin(), (x).end()

const ll INF = 100100100100100100;
const ll MOD = 1000000007;

// get abs
ll my_abs(ll a);
// a^n
ll a_n(ll a, ll n);
// get gcd
ll my_gcd(ll a, ll b);
// a^(-1) % MOD 
ll inv(ll a);

int main() {
    ll n, m, c;
    cin >> n >> m >> c;
    vector<ll> b(m);
    rep(i, m) cin >> b[i];
    ll ans = 0;
    rep(i, n) {
        ll tmp = 0;
        rep(j, m) {
            ll a;
            cin >> a;
            tmp += b[j]*a;
        }
        if(tmp+c>0) ans++;
    }
    cout << ans << endl;
    return 0;
}

ll my_abs(ll a) {
    if(a >= 0) return a;
    else return -1 *a;
}

ll a_n(ll a, ll n) { 
    if(n == 0) return 1;
    
    ll ret = a, count = 1;
    while(count * 2 < n) {
        ret *= ret;
        count *= 2;
    }
    if(count == n) return ret;
    else return (ret * a_n(a, n-count));
}

ll my_gcd(ll a, ll b) {
    if(b == 0) return a;
    return my_gcd(b, a%b);
}

ll inv(ll a) {
    return a_n(a, MOD-2);
}