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 (GCC 5.4.1) (judge verdict: Accepted)

char buf[30000];
sz;
N,H,W;
main()
{
	read(0,buf,30000);
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		N=N*10+k-'0';
	}
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		H=H*10+k-'0';
	}
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		W=W*10+k-'0';
	}
	int ans=0;
	for(int i=0;i<N;i++)
	{
		int A=0,B=0;
		for(;;)
		{
			char k=buf[sz++];
			if(k<'0')break;
			A=A*10+k-'0';
		}
		for(;;)
		{
			char k=buf[sz++];
			if(k<'0')break;
			B=B*10+k-'0';
		}
		ans+=A>=H&&B>=W;
	}
	printf("%d\n",ans);
}