The purpose of this problem is to verify whether the method you are using to read input data is sufficiently fast to handle problems branded with the enormous Input/Output warning. You are expected to be able to process at least 2.5MB of input data per second at runtime.

Input

The input begins with two positive integers n k (n, k<=10⁷). The next n lines of input contain one positive integer t_i, not greater than 10⁹, each.

Output

Write a single integer to output, denoting how many integers t_i are divisible by k.

题解：记录这道题主要是为了记录java中Scanner和BufferReader的区别，开始用Scanner，效率非常低，所以就TLE了。根据StackOverFlow里面解释：BufferReader只是从流中读入数据，但不对数据做任何处理，Scanner按照需求解析数据并读入，比如nextInt(),nextDouble()等。更详细的答案还有这里。

总结一下：

• A BufferedReader is a simple class meant to efficiently read from the underling stream.
•  BufferedReader is synchronized, so read operations on a BufferedReader can safely be done from multiple threads.
• Scanner can parse the underlying stream for primitive types and strings using regular expressions.
• A scanner however is not thread safe, it has to be externally synchronized.

对于原文中的“ A scanner can do all that a BufferedReader can do and at the same level of efficiency as well.”不太认同，因为通过OJ来看，BufferReader的效率确实比Scanner高。

BufferReader的用法就用这道题的AC代码记录：

 1 import java.io.BufferedReader; 2 import java.io.IOException; 3 import java.io.InputStreamReader; 4  5 public class Main { 6  7     public static void main(String[] args) { 8         // TODO Auto-generated method stub 9         BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));10         try{11             String[] line =(bf.readLine()).split(" ");12             int n = Integer.parseInt(line[0]);13             int k = Integer.parseInt(line[1]);14             int count = 0;15             while(n-- > 0){16                 int num = Integer.parseInt(bf.readLine());17                 if(num%k == 0)18                     ++count;19             }20             System.out.println(count);21         }22         catch(IOException e){23             System.out.print("input error");24         }25     }26 27 }