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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define N_MAX 20
#define M_MAX 20

int main(){

    int N, M;
    int A[N_MAX][M_MAX] = {0};
    int B[M_MAX] = {0};
    int C;

    int i, j, k;
    int sum = 0;
    int cnt = 0;

    // 入力
    scanf("%d %d %d", &N, &M, &C);
  	//printf("%d %d %d\n", N, M, C);
    for(i=0 ; i<M ; i++){
        scanf("%d", &B[i]);
        //printf("%d ", B[i]);
    }
    //printf("\n");
    for(i=0 ; i<N ; i++){
        for(j=0 ; j<M ; j++){
            scanf("%d", &A[i][j]);
            //printf("%d ", A[i][j]);
        }
    //printf("\n");
    }

    for(i=0 ; i<N ; i++){
        sum = 0;
        for(j=0 ; j<M ; j++){
            sum += (A[i][j]*B[j]);
        }
        sum += C;
        if(sum > 0){
            cnt++;
        }
    }

    printf("%d", cnt);
}