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>

int main(void){
    int N,M,C;
    scanf("%d %d %d",&N,&M,&C);
    int B[M],A[N][M];
    for(int i=0;i<M;i++)
    scanf("%d",&B[i]);
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            scanf("%d",&A[i][j]);
        }
    }    

    //////////////*確認用*///////////////////
    /*
    printf("------B行列------\n");
    for(int i=0;i<M;i++)
    printf("%d ",B[i]);
    printf("\n");
    printf("------A行列------\n");
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            printf("%d ",A[i][j]);
        }
        printf("\n");
    }
    */
    /////////////////////////////////////////
    int count=0;
    for(int i=0;i<N;i++){
        int sum=0;
        for(int j=0;j<M;j++){
            sum=sum+B[j]*A[i][j];
        }        
        sum=sum+C;
        //printf("sumは%d\n",sum);

        if(sum>0){
        count+=1;
        }
    }

    printf("%d\n",count);
}