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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func read() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	v, _ := strconv.Atoi(read())
	return v
}

func main() {
	sc.Split(bufio.ScanWords)

	N := nextInt()
	M := nextInt()
	C := nextInt()

	// B := make([]int, M)
	var B [20]int
	for i := 0; i < M; i++ {
		B[i] = nextInt()
	}
	// fmt.Println(B)

	// A := make([][]int, N)
	var A [20][20]int
	for i := 0; i < N; i++ {
		// A[i] = make([]int, M)
		for j := 0; j < M; j++ {
			A[i][j] = nextInt()
		}
	}

	// fmt.Println(A)
	var sum, count int

	for i := 0; i < N; i++ {
		sum = 0
		for j := 0; j < M; j++ {
			sum += A[i][j] * B[j]
		}
		fmt.Println(sum)
		if sum+C > 0 {
			count++
		}
	}
	fmt.Println(count)

}