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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int (i) = 0; (i) < (n); ++(i))
#define rrep(i,n) for(int (i) = (n) - 1; (i) >= 0; --(i)) //Reversi rep
#define nfor(i,a,b) for(int (i) = (a); (i) < (b); ++(i)) //Natural for
#define rfor(i,a,b) for(int (i) = (b) - 1; (i) >= (a); --(i)) //Reversi for
#define ll long long
#define mod (1000000007)
#define YES cout << "YES" << endl;
#define Yes cout << "Yes" << endl;
#define yes cout << "yes" << endl;
#define NO cout << "NO" << endl;
#define No cout << "No" << endl;
#define no cout << "no" << endl;

int asc(const void *a, const void *b) {
    return *(int *)a - *(int *)b;
}

int desc(const void *a, const void *b) {
    return *(int *)b - *(int *)a;
}

int main()
{
    ll n,h,w,a,b,ans = 0;
    cin >> n >> h >> w;
    rep(i, n){
        cin >> a >> b;
        if (a >= h && b >= w) {
            ++ans;
        }
    }
    cout << ans << endl;
    return 0;
}