#include<bits/stdc++.h>
//#define int long long
#define ll long long
#define ld long double
#define _1 first
#define _2 second
#define yes cout<<"Yes\n"
#define nah cout<<"No\n"
#define FFF ios_base::sync_with_stdio(0);cin.tie(0);
#define ipr pair<int,int>
#define ret return
#define intt int32_t
#define mid ((l+r)/2)
#define pb push_back
#define lll __int128
using namespace std;
int tst, ts;
intt mo = 0*(998244353)+(1e9+7)*1, dx[] = {0, -1, -1, 0}, dy[] = {1, 1, 2, 0};
int mul( int x, int y ) {
ret (int) ( (ll) x%mo * y%mo % mo);
ret x*y;
}
int pwo( int x, ll y ) {
int res = 1;
for( int i = 63; i + 1; i-- )
res = mul( res, mul(res , ( ( 1ll << i )&y ? x : 1 )) );
ret res;
}
int dvii( int x, int y ) {
ret mul( x, pwo(y,mo-2) );
}
int oo( char x , char y) {
ret ( int )x - y;
}
int lgg( int x, int y ) {
int u = 0;
while( x ) {
u++;
x /= y;
}
ret u;
}//g++ -Wall -o "%e" "%f" && "./%e"
int mun( int x, int y ) {
while( x < y )x += mo;
ret ( x - y ) % mo;
}
int add( int x, int y ) {
ret x + y - ( mo * ( x + y >= mo ) );
ret x + y;
}
int lcm( int x, int y ) {
if(x*y==0)ret max(x,y);
int o=__gcd(x,y);
x/=o;
ret x*y;
}
#define endl '\n';
#define nl no*2
#define nr no*2+1
const int M =1e7+7, N = 507, P = 1e12, inf = 1e9+7 ;
int n,k,l[N],ifa[N],cn[N*4][N],cnn[N][N],ccn[N*4][N],dp[2][N],r[N],c[N*4];
vector<int>vv[N*4];
void solve(){
cin>>n;
set<int>st;
vector<int>v(1,inf);
for(int i=1;i<=n;i++){
cin>>l[i]>>r[i];
st.insert(l[i]);
st.insert(r[i]);
}
for(auto i:st){
if(v.back()<i-1){
c[v.size()]=i-v.back()-1;
v.pb(i-1);
}
c[v.size()]=1;
v.pb(i);
}
/// for every i:0 < i < v.size()-1 the points from v[i-1]+1 to v[i+1]-1 have the same set of intervals
int ii=1;
for(int i=0;i<=n;i++){
ii=mul(ii,max(i,1));
for(int j=0;j<=i;j++)
cnn[i][j]=mul(ii,mul(ifa[j],ifa[i-j]));///cnn[i][j]=cnk(i,j)
}
for(int i=1;i<v.size();i++){
ii=1;
for(int j=0;j<=min(c[i],n);j++){
ccn[i][j]=mul(ii,ifa[j]);///ccn[i][j]=cnk(c[i],j)
ii=mul(ii,c[i]-j);
}
for(int j=1;j<=n;j++)
if(l[j]<=v[i]&&v[i]<=r[j])
vv[i].pb(j);
}
for(int i=1;i<v.size();i++)
for(int j=2;j<=n && c[i]>1;j++)
for(int p=0;p<=min(c[i],j)-2;p++){
cn[i][j]=add(cn[i][j],mul(cnn[j-2][p],ccn[i][p+2]));
/// p+2 is number of intervals have been chosen whith some points from v[i-1]+1 to v[i+1]-1
/// j is maximum interval has been chosen
/// c[i] the number of points from v[i-1]+1 to v[i+1]-1 ==>c[i]=v[i+1]-v[i-1]-1
}
dp[0][0]=1;
///dp[i][j] the number of ways to be j is maximum interval has been chosen and the point v[i] is maximum points may be chosen
for(int i=1;i<v.size();i++){
int o=0,kk=0;
for(int j=0;j<=n;j++)
dp[i%2][j]=dp[(i-1)%2][j];/// to don't chose any thing
for(int j=1;j<=n;j++){
o=add(o,dp[(i-1)%2][j-1]);
if(l[j]>v[i]||v[i]>r[j])
continue;
while(kk<vv[i].size()&&vv[i][kk]<=j)kk++;
dp[i%2][j]=add(dp[i%2][j],mul(c[i],o));/// to chose just the interval j
for(int k=kk;k<vv[i].size()&&c[i]>1;k++){
/// j is minimum interval has been chosen
/// vv[i][k] is maximum interval has been chosen
dp[i%2][vv[i][k]]=add(dp[i%2][vv[i][k]],mul(o,cn[i][k-kk+2]));
}
}
}
int o=0;
for(int i=1;i<=n;i++)
o=add(o,dp[(v.size()-1)%2][i]);
cout<<o<<endl;
}
intt main() {
FFF
ifa[0]=1;
for(int i=1;i<N;i++)
ifa[i]=mul(dvii(1,i),ifa[i-1]);
//freopen("7.in", "r", stdin);
//freopen("out.txt", "w", stdout);
tst = 1;
//cin >> tst;
for( ts = 1; ts <= tst; ts++ ){
solve();
}
}
