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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef  long long ll;
//double pi=3.1415926535898;
//double pi=acos(-1.0);
#define scarr(a,s,e) for(int i=s;i<=int(e);i++) scanf("%d ",&a[i]);
#define prarr(a,s,e) for(int i=s;i<=int(e);i++) printf("%d ",a[i]);
#define speed      ios::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
#define rng(i,a,b) for(int i=a;i>=int(b);i--)
#define rep(i,a,b) for(int i=a;i<=int(b);++i)
#define lop(i,n)   for(int i=0;i<int(n);i++)
#define ms(x,a)    memset(x,a,sizeof(x))
#define all(x)     (x).begin(),(x).end()
#define sz(x)      int(x.size())
#define pb         push_back
#define mkp        make_pair
#define fr         first
#define se         second
#define sc(x)      scanf("%d",&x)
#define pr(x)      printf("%d ",x)
#define scl(x)     scanf("%lld",&x)
#define prl(x)     printf("%lld ",x)
#define prln(x)    printf("%d\n",x)
#define r0         return 0;
#define CC         continue;
#define scd(x)     scanf("%lf",&x)
#define prd(x)     printf("%.10f\n",x);
#define bb(x)      cerr<< x <<endl;
#define ii         pair<int,int>
//#define max(a,b) a<b?b:a
//#define min(a,b) a<b?a:b
const ll M   = 1e7+9 ;
const ll inf = 1e19  ;
const int si = 1e5+9 ;
//////////////////////    Hi    ^_^  hack me if you can :P

int main()
{
    int n,a,b,p;
    int ans = 0;
    cin>>n>>a>>b;
    lop(i,n){
    cin>>p;
    if(p<a||p>=b){
        ++ans;
    }
    }
   cout<<ans;

    r0;
}


/*

ll pwmod(ll a, ll n, ll mod) {
    ll ret = 1;
    while (n) {
        if (n & 1) ret = ret * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return ret;
}

*/
/*
freopen("output.txt","w",stdout);
freopen("input.txt","r",stdin);
for (int i=0;i<n;i++)            Hi    ^_^  hack me if you can :P
int mx[]= {0,0,1,1,1,-1,-1,-1};
int my[]= {1,-1,0,1,-1,0,1,-1};
bool inside (int i,int j) {return (i>=0 && i<n && j>=0 && j<m);}
*/