Snuke arranged N colorful balls in a row. The i-th ball from the left has color c_i and weight w_i.

He can rearrange the balls by performing the following two operations any number of times, in any order:

* Operation 1: Select two balls with the same color. If the total weight of these balls is at most X, swap the positions of these balls.
* Operation 2: Select two balls with different colors. If the total weight of these balls is at most Y, swap the positions of these balls.



How many different sequences of colors of balls can be obtained? Find the count modulo 10^9 + 7.

Constraints

* 1 ≤ N ≤ 2 × 10^5
* 1 ≤ X, Y ≤ 10^9
* 1 ≤ c_i ≤ N
* 1 ≤ w_i ≤ 10^9
* X, Y, c_i, w_i are all integers.

Input

Input is given from Standard Input in the following format:


N X Y
c_1 w_1
:
c_N w_N


Output

Print the answer.

Examples

Input

4 7 3
3 2
4 3
2 1
4 4


Output

2


Input

1 1 1
1 1


Output

1


Input

21 77 68
16 73
16 99
19 66
2 87
2 16
7 17
10 36
10 68
2 38
10 74
13 55
21 21
3 7
12 41
13 88
18 6
2 12
13 87
1 9
2 27
13 15


Output

129729600
#include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
const int mod=1e9+7,N=200005,inf=0x3f3f3f3f;
vector<int> orz[N];
int mi=inf,mic,mii,X,Y,ans,js,n,a[N],fac[N],ni[N];
int ksm(int x,int y) {
	int re=1;
	for(;y;y>>=1,x=1LL*x*x%mod) if(y&1) re=1LL*re*x%mod;
	return re;
}
int main()
{
	int c,w;
	n=read(),X=read(),Y=read();
	for(RI i=1;i<=n;++i) {
		c=read(),w=read(),orz[c].push_back(w);
		if(w<=mi) mii=mi,mi=w,mic=c;
		else if(w<mii) mii=w;
	}
	for(RI i=1;i<=n;++i) sort(orz[i].begin(),orz[i].end());
	for(RI i=1;i<=n;++i) {
		if(!(int)(orz[i].size())) continue;
		if(i==mic) {
			a[i]=1,++js;
			for(RI j=1;j<orz[i].size();++j)
				if(orz[i][j]+mii<=Y||orz[i][j]+orz[i][0]<=X) ++a[i],++js;
		}
		else {
			if(orz[i][0]+mi>Y) continue;
			a[i]=1,++js;
			for(RI j=1;j<orz[i].size();++j)
				if(orz[i][j]+mi<=Y||orz[i][j]+orz[i][0]<=X) ++a[i],++js;
		}
	}
	fac[0]=1;for(RI i=1;i<=n;++i) fac[i]=1LL*fac[i-1]*i%mod;
	ni[n]=ksm(fac[n],mod-2);
	for(RI i=n-1;i>=0;--i) ni[i]=1LL*ni[i+1]*(i+1)%mod;
	ans=fac[js];
	for(RI i=1;i<=n;++i) ans=1LL*ans*ni[a[i]]%mod;
	printf("%d\n",ans);
    return 0;
}