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: Wrong Answer)

#include<stdio.h>

int main(void){
    int n,m,c = 0;
    int d = 0;
    int h, sum = 0;
    scanf("%d %d %d", &n, &m, &c);

    int b[m];
    int a[n][m];
    for(int i=0;i<m;i++){
        scanf("%d", &b[i]);
    }

    //output
    /*for(int i=0;i<m;i++){
        printf("%d ", b[i]);
    }
    printf("\n\n");
    */

    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            scanf("%d", &a[i][j]);
        }
    }

    /*//output
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            printf("%d ", a[i][j]);
        }
        printf("\n\n");
    }
    */

    for(int i=0;i<n;i++){
        sum = 0;
        for(int j=0;j<m;j++){
            sum += a[i][j] * b[j];
            if(j == m-1){
                //printf("%d\n", sum);
                sum += c;
                if(sum > 0)h++;
            }
        }
    }
    printf("%d\n", h);
}