Problem:
Score : 66 points

Problem Statement

Summer vacation ended at last and the second semester has begun.
You, a Kyoto University student, came to university and heard a rumor that somebody will barricade the entrance of your classroom.
The barricade will be built just before the start of the A-th class and removed by Kyoto University students just before the start of the B-th class.
All the classes conducted when the barricade is blocking the entrance will be cancelled and you will not be able to attend them.
Today you take N classes and class i is conducted in the t_i-th period.
You take at most one class in each period.
Find the number of classes you can attend.

Constraints

1 \leq N \leq 1000

1 \leq A < B \leq 10^9

1 \leq t_i \leq 10^9

All t_i values are distinct.

Input

N, A and B are given on the first line and t_i is given on the (i+1)-th line.

N A B
t1
:
tN

Output

Print the number of classes you can attend.

Sample Input 1

5 5 9
4
3
6
9
1

Sample Output 1

4

You can not only attend the third class.

Sample Input 2

5 4 9
5
6
7
8
9

Sample Output 2

1

You can only attend the fifth class.

Sample Input 3

4 3 6
9
6
8
1

Sample Output 3

4

You can attend all the classes.

Sample Input 4

2 1 2
1
2

Sample Output 4

1

You can not attend the first class, but can attend the second.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define miin(a, b, c) min((a), min((b), (c)))
#define maax(a, b, c) max((a), max((b), (c)))
#define mod 1000000007
#define INF 2002002002LL
const int NIL = -1;

int n,a,b,c=0,t;

main(){
    scanf("%d%d%d", &n, &a, &b);
    for(int i=0; i<n; i++){
        scanf("%d", &t);
        if(a <= t && t < b)c++;
    }
    printf("%d\n", n - c);
    return 0;
}