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>

char word[10+1];
char line[11+2];
char end[] = "END_OF_TEXT";

int equals(char *a, char *b) {
    int i = 0;
    while(1) {
        if(a[i] != b[i] && (a[i]-b[i])*(a[i]-b[i]) != ('a'-'A')*('a'-'A')) return 0;
        else if(a[i] == '\0') return 1;
        i++;
    }
}

int is_end(char *t) {
//    printf("owari!\n");
    return equals(t,end);
}

int main(void) {

    int i = 0;
    while(1) {
        char c;
        scanf("%c", &c);
        if(c == ' ' || c == '\n') break;
        word[i++] = c;
    }
    word[i] = '\0';

    int count = 0;
    while(1) {
        i = 0;
        while(1) {
            if(i>11) break;
            char c;
            scanf("%c", &c);
            if(c == ' ' || c == '\n') break;
            line[i++] = c;
        }
        line[i] = '\0';
//        printf("%s\n", line);
        if(is_end(line)) break;
        if(equals(word, line)) count++;
    }

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