Problem:
Score : 200 points

Problem Statement

There are N rectangular plate materials made of special metal called AtCoder Alloy.
The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally).

Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W.
He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary.
When cutting a material, the cuts must be parallel to one of the sides of the material.
Also, the materials have fixed directions and cannot be rotated.
For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate.

Out of the N materials, how many can produce an H \times W plate if properly cut?

Constraints

1 \leq N \leq 1000

1 \leq H \leq 10^9

1 \leq W \leq 10^9

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N H W
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the answer.

Sample Input 1

3 5 2
10 3
5 2
2 5

Sample Output 1

2

Takahashi wants a 5 \times 2 plate.

The dimensions of the first material are 10 \times 3. We can obtain a 5 \times 2 plate by properly cutting it.

The dimensions of the second material are 5 \times 2. We can obtain a 5 \times 2 plate without cutting it.

The dimensions of the third material are 2 \times 5. We cannot obtain a 5 \times 2 plate, whatever cuts are made. Note that the material cannot be rotated and used as a 5 \times 2 plate.

Sample Input 2

10 587586158 185430194
894597290 708587790
680395892 306946994
590262034 785368612
922328576 106880540
847058850 326169610
936315062 193149191
702035777 223363392
11672949 146832978
779291680 334178158
615808191 701464268

Sample Output 2

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<iostream>
#include<vector>
#include<cstdlib>
using namespace std;
 
template <class T> using V = vector<T>;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
const long long INF = 1LL << 60;
 
using ll = long long;
using db = double;
using st = string;
using ch = char;
using bl = bool;
using vll = V<ll>;
using vpll =V<pair<ll,ll>>;
using vst = V<st>;
using vdb = V<db>;
using vch = V<ch>;
using vbl = V<bl>;
 
 
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rFOR(i,a,b) for(ll i=(a);i>(b);i--)
#define oFOR(i,a,b) for(ll i=(a);i<(b);i+=2)
#define bgn begin()
#define en end()
#define SORT(a) sort((a).bgn,(a).en)
#define REV(a) reverse((a).bgn,(a).en)
#define M(a,b) max(a,b)
#define rM(a,b) min(a,b)
#define fi first
#define se second
#define sz size()
#define gcd(a,b) __gcd(a,b)
#define co(a) cout<<a<<endl;
#define ci(a) cin>>a;
 
 
ll sum(ll n) {
  ll m=0;
  FOR(i,0,20){
    m+=n%10;
    n/=10;
    if(n==0){
      break;
    }
  }
  return m;
}
 
 
const int MAX = 510000;
const int MOD = 1000000007;
 
long long fac[MAX], finv[MAX], inv[MAX];
 
void Comuse() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
 
#define comuse Comuse()
 
ll combi(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
 
 
 
ll lcm(ll a,ll b){
  ll n;
  n=a/gcd(a,b)*b;
  return n;
}
 
 
/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/
 
 
signed main() {
  ll n;
  	ci(n);
  ll h,w;
  	ci(h>>w);
  ll ans=0;
  FOR(i,0,n){
    ll a,b;
    	ci(a>>b);
    if(a<=h&&b<=w){
      ans++;
    }
  }
  co(ans)
}