#include<bits/stdc++.h>
#define mp make_pair
#define fs first
#define sc second
#define pb push_back
#define debug(x) cout<<#x<<" = "<<(x)<<endl
#define mod 1000000007
#define INF 1e18
using namespace std;
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
int n;
pair<int,int> tim[1000005];
int parent[2000005];
vector<int> tree[4 * 1000005];
int caribapak(int now){
if(parent[now] == now) return now;
return parent[now] = caribapak(parent[now]);
}
void gabung(int a,int b){
int x = caribapak(a);
int y = caribapak(b);
if(x != y){
parent[x] = y;
return;
}
}
void update(int idx,int l,int r,int pos,int val){
if(pos > r || pos < l){
return;
}
tree[idx].pb(val);
if(l == r){
return;
}
update(2 * idx, l, (l + r)/2, pos, val);
update(2 * idx + 1, (l + r)/2 + 1, r, pos, val);
}
void query(int idx,int l,int r,int kiri,int kanan,int val){
if(kiri > r || kanan < l){
return;
}
if(l >= kiri && r <= kanan){
for(int i = 0; i < tree[idx].size(); i++){
int now = tree[idx][i];
// if(val == 4) cout << now << " tes\n";
int com = now + n;
if(now > n) com = now - n;
gabung(now, val + n);
gabung(com, val);
}
if(tree[idx].size() > 0){
tree[idx].clear();
tree[idx].pb(val + n);
}
return;
}
query(2 * idx, l, (l + r)/2, kiri, kanan, val);
query(2 * idx + 1, (l + r)/2 + 1, r, kiri, kanan, val);
}
int main(){
cin >> n;
for(int i = 1 ; i <= n; i++){
cin >> tim[i].fs >> tim[i].sc;
}
sort(tim + 1, tim + n + 1);
for(int i = 1; i <= 2 * n; i++){
parent[i] = i;
}
for(int i = 1; i <= n; i++){
query(1, 1, 2 * n, tim[i].fs, tim[i].sc, i);
update(1, 1, 2 * n, tim[i].sc, i);
}
for(int i = 1; i <= n; i++){
int x = caribapak(i);
int y = caribapak(i + n);
if(x == y){
// cout << i << " haha\n";
cout << "0\n";
return 0;
}
}
long long ans = 1;
int nyak = 0;
for(int i = 1; i <= 2 * n; i++){
if(caribapak(i) == i){
nyak++;
// cout << i << " here\n";
// ans = (ans * 2) % mod;
}
}
for(int i = 0; i < nyak / 2; i++){
ans = (ans * 2) % mod;
}
cout << ans << "\n";
}
