Problem:
Finding a Word

Write a program which reads a word W and a text T, and prints the number of word W which appears in text T

T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.

Constraints

The length of W ≤ 10

W consists of lower case letters

The length of T in a line ≤ 1000

Input

In the first line, the word W is given. In the following lines, the text T is given separated by space characters and newlines.

"END_OF_TEXT" indicates the end of the text.

Output

Print the number of W in the text.

Sample Input

computer
Nurtures computer scientists and highly-skilled computer engineers
who will create and exploit "knowledge" for the new era.
Provides an outstanding computer environment.
END_OF_TEXT

Sample Output

3

Solution in C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
int main(){
  char line[1000];
  char word[12];
  char end[]="END_OF_TEXT";		/* 終了条件 */
  char end2[]="END_OF_TEXT\n";
  int count=0;					/* lineの中のwordの数 */
  char tmp[10];
  int i;
  int length;
  //wordの取得
  fgets(word, sizeof(word), stdin);
  word[strlen(word)-1]='\0';
  //  printf("word : %s\n", word);
  //チェック
  char *str_ptr;
  while(1){
	fgets(line, sizeof(line), stdin);
	line[strlen(line)-1]='\0';	/* AOJに提出するときはこのコードは消す */

	if((strcmp(line, end)==0)||(strcmp(line, end2)==0))	/* 終了条件 */
	  break;

	str_ptr=strtok(line, " ");
	/* printf("done\n"); */
	/* fflush(stdout); */
	
	while(1){
	  if(str_ptr==NULL)
		break;
	  if(strcasecmp(word, str_ptr)==0)
		++count;
	  str_ptr=strtok(NULL, " "); 
	}
  }
  printf("%d\n", count);
  return 0;
}