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 <iostream>
#include<algorithm>
#include<string>
#include <cmath>
#include <vector>
#include <map>
#include <cstdio>
#include <iomanip>
#pragma region Macros
#define int long long
#define double long double
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
#define LAST(x) x[x.size()-1]
#define ALL(x) (x).begin(),(x).end()
#define swap(a,b) (a += b,b = a - b,a -= b)
#define CEIL(a,b) ((a+b-1)/b)
int FACT(int a) {
	if (a == 0)
		return 1;
	else
		return a * FACT(a - 1);
}
int nPr(int n, int r) {
	int s = n - r + 1;
	int sum = 1;
	for (int i = s; i <= n; i++)
		sum *= i;
	return sum;
}
//int nCr(int n, int r)
int nCr2(int n, int r) {
	return FACT(n) / (FACT(r) * FACT(n - r));
}
int GCD(int a, int b) // assuming a,b >= 1
{
	if (a < b)
		swap(a, b);
	if (b == 0)
		return a;
	if (a % b == 0)
		return b;
	return GCD(b, a % b);
}
int LCM(int a, int b) // assuming a,b >= 1
{
	return  a * b / GCD(a, b);
}
double LOG(int a, int b) {
	return log(b) / log(a);
}
inline bool BETWEEN(int x, int min, int max) {
	if (min <= x && x <= max)
		return true;
	else
		return false;
}
inline bool PRIMEOR (int x) {
	if (x ==1)
		return false;
	if (x == 2)
		return true;
	if (x%2==0)
		return false;
	double sqrtx = sqrt(x);
	for (int i = 3; i <= sqrtx; i += 2) {
		if (x % i == 0)
			return false;
	}
	return true;
}

using namespace std;
#pragma endregion
signed main() {
	int n,h,w, x = 0,A,B;
	cin >> n >> h >> w;
	rep(i, n) {
		cin >> A >> B;
		if (A >= h && B >= w)
			x++;
	}
	cout << x <<endl;
}