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 <stdlib.h>
int main(void){
	int i,len,kensaku=0,ENDcount=0;
	char str[1200],W[10],T[1000],serchlen,Wcount=0,END[]="END_OF_TEXT";
	fgets(W,sizeof W,stdin);
	serchlen=strlen(W);
	for(i=0;i<serchlen-1;i++){//serch word input
			if( isupper(T[i]) ){
				T[i]=tolower(T[i]);//omoji->komoji
			}
		}
	while(1){
		fgets(T,sizeof T,stdin);//text input
		len=strlen(T);
		for(i=0;i<len-1;i++){
			if(T[i]==END[i]){
				ENDcount++;
			}
			else{
				ENDcount=0;
				}
			if(ENDcount==11){
				goto OUT;//while nukeru
			}
		}
		for(i=0;i<len-1;i++){
			if( isupper(T[i]) ){
				T[i]=tolower(T[i]);//omoji->komoji
			}
		}
		for(i=0;i<len-1;i++){
			if(T[i]==W[kensaku]){
				kensaku++;
				if(kensaku==serchlen-1){
					Wcount++;
					kensaku=0;
				}
			}
			else{
				kensaku=0;
			}
		}
		
	}
	OUT:
	printf("%d\n",Wcount);
	
	
	return 0;
}