#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 1000001;
int N;
pi par[MX];
vi rap[MX][2];
map<int,int> M[MX];
vpi p;
map<int,int> m;
void input() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N; p.resize(N);
m.clear();
F0R(i,N) {
F0R(j,2) rap[i][j].clear();
M[i].clear();
}
F0R(i,N) {
cin >> p[i].f >> p[i].s;
par[i] = {i,0};
rap[i][0].pb(i);
M[i][p[i].s] = i;
}
sort(all(p));
}
void unite(int a, int b) {
// cout << "AH " << a << " " << b << "\n";
if (par[a].f == par[b].f) {
if (par[a].s == par[b].s) {
cout << 0;
exit(0);
}
return;
}
bool eq = (par[a].s == par[b].s);
a = par[a].f, b = par[b].f;
if (sz(rap[a][0])+sz(rap[a][1]) < sz(rap[b][0])+sz(rap[b][1])) swap(a,b);
F0R(j,2) for (int i: rap[b][j]) {
par[i] = {a,j^eq};
rap[a][j^eq].pb(i);
M[a][p[i].s] = i;
}
F0R(j,2) rap[b][j].clear();
M[b].clear();
}
void del(int mn) {
int x = par[m.begin()->s].f;
m.erase(m.begin());
auto IT = M[x].lb(mn);
if (IT == M[x].end()) return;
m[IT->f] = IT->s;
}
void check(vi v) {
sort(all(v));
/*for (int i: v) cout << i << " * ";
cout << "BAD\n";
for (int i: v) cout << p[i].f << " " << p[i].s << " | ";
cout << "\n";*/
set<int> z;
for (int i: v) {
while (sz(z) && *z.begin() < p[i].f) z.erase(z.begin());
if (sz(z) && *z.begin() < p[i].s) {
// cout << "\nAH " << i << " " << p[i].f << " " << p[i].s << " " << *z.begin() << " " << sz(z) << "\n";
cout << 0;
exit(0);
}
z.insert(p[i].s);
}
}
void solve() {
input();
F0R(i,sz(p)) {
while (sz(m) && m.begin()->f < p[i].f) del(p[i].f);
vi to;
for (auto a: m) {
if (a.f < p[i].s) to.pb(a.s);
else break;
}
for (int j: to) {
m.erase(p[j].s);
unite(j,i);
}
to.pb(i);
m[p[to.front()].s] = to.front();
}
int ans = 1;
F0R(i,N) if (par[i].f == i) {
ans = 2*ans%MOD;
// check(rap[i][0]);
// check(rap[i][1]);
}
cout << ans << "\n";
}
int main() {
F0R(i,1) solve();
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
