#include <bits/stdc++.h>
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<pll, ll> edge;
typedef pair<int, int> pii;
const int Maxn = 1e3 + 10;
const int Inf = 1e9;
const int Log = 20;
const int Mod = 1e9 + 7;
int mul(int a, int b){
return (1ll * a * b) % Mod;
}
int pw(int b, int p){
if(p == 0) return 1;
if(p & 1) return mul(b, pw(b, p - 1));
return pw(mul(b, b), p >> 1);
}
int inv[Maxn * 4];
int a[Maxn], b[Maxn];
set<int> st;
vector<pii> V;
int dp[2][1040][510];
int sm[2][1040];
int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
for(int i = 0; i < 4 * Maxn; i++) inv[i] = pw(i, Mod - 2);
int n;
cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i] >> b[i];
b[i] ++;
st.insert(a[i]);
st.insert(b[i]);
//assert(a[i] == b[i] - 1);
}
ll l = -1;
for(auto x : st){
if(l != -1) V.pb({l, x});
l = x;
}
sort(all(V));
int m = V.size();
for(int I = 1; I <= n; I++){
int II = I & 1;
memset(dp[II], 0, sizeof dp[II]);
memset(sm[II], 0, sizeof sm[II]);
for(int i = 0; i < m; i++){
if( (a[I] <= V[i].F) && (V[i].S <= b[I]) ){
for(int cnt = 2; cnt <= I; cnt ++){
//if(cnt > V[i].S - V[i].F) continue;
dp[II][i][cnt] = mul( dp[II ^ 1][i][cnt - 1], mul((V[i].S - V[i].F) - cnt + 1, inv[cnt]) );
}
}
if( (a[I] <= V[i].F) && (V[i].S <= b[I]) ){
for(int j = 0; j < i; j++){
dp[II][i][1] += mul( sm[II ^ 1][j], V[i].S - V[i].F );
if(dp[II][i][1] >= Mod)
dp[II][i][1] %= Mod;
}
}
}
for(int i = 0; i < m; i++){
if( (a[I] <= V[i].F) && (V[i].S <= b[I]) ){
dp[II][i][1] += (V[i].S - V[i].F);
//S += (V[i].S - V[i].F);
dp[II][i][1] %= Mod;
}
}
for(int i = 0; i < m; i++) for(int cnt = 1; cnt <= I; cnt ++){
dp[II][i][cnt] += dp[II ^ 1][i][cnt];
if(dp[II][i][cnt] >= Mod)
dp[II][i][cnt] -= Mod;
}
for(int i = 0; i < m; i++) for(int cnt = 1; cnt <= I; cnt ++){
sm[II][i] += dp[II][i][cnt];
//cerr << I << " " << dp[II][i][cnt] << "\n";
//sm[II][i] %= Mod;
if(sm[II][i] >= Mod)
sm[II][i] -= Mod;
}
}
ll ans = 0;
for(int i = 0; i < m; i++) ans += sm[n & 1][i];
cout << ((ans % Mod) + Mod) % Mod << '\n';
return 0;
}
