#include<bits/stdc++.h>
using namespace std;
#define int long long
#define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
#define FORD(i, a, b) for (int i = (a), _b = (b); i >= _b; i--)
#define fi first
#define se second
#define pb push_back
#define ALL(a) (a).begin(), (a).end()
#define task "kbsiudthw"
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<int, ii> pii;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, L[505], R[505], dp[505][5005], pref[505][5005], dp2[5005][505], inv[505];
int Pow(int a, int b){
if (b == 0) return 1;
int t = Pow(a, b/2);
t = (t * t)%MOD;
if (b%2 == 0) return t;
return (t * a)%MOD;
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
if (fopen(task".inp", "r")){
freopen(task".inp", "r", stdin);
freopen(task".out", "w", stdout);
}
cin >> n ;
FOR (i, 1, n) {
cin >> L[i] >> R[i];
}
map<int, int> dd;
vi vals;
FOR (i, 1, n){
int x = R[i];
vals.pb(x);
vals.pb(L[i]);
if (x - 1) vals.pb(x-1);
vals.pb(x+1);
}
sort(ALL(vals));
vals.resize(unique(ALL(vals)) - vals.begin());
int m = (int)vals.size();
FOR (i, 1, n){
L[i] = upper_bound(ALL(vals), L[i]) - vals.begin();
R[i] = upper_bound(ALL(vals), R[i]) - vals.begin();
}
FOR (i, 1, 504) inv[i] = Pow(i, MOD - 2);
dp[0][m + 1] = 1;
FOR (i, 1, n){
FORD(j, i-1, 0){
FOR (k, L[i], R[i] - 1){
int cnt = vals[k]-vals[k-1];
dp[i][k] = (dp[i][k] + (pref[j][k-1] * cnt)%MOD)%MOD;
if (k >= R[j]){
dp[i][k] = (dp[i][k] + (dp[j][m+1] * (cnt - (k == R[j])))%MOD)%MOD;
}
}
dp[i][m+1] = (dp[i][m+1] + pref[j][R[i]-1])%MOD;
if (R[i] > R[j]){
dp[i][m+1] = (dp[i][m+1] + dp[j][m+1])%MOD;
}
}
FOR (k, L[i], R[i] - 1){
int cnt = vals[k] - vals[k-1];
int ans = 0;
FORD(j, min(n, cnt), 2){
int t = inv[j] * (cnt - j + 1)%MOD;
ans = (ans + dp2[k][j-1] * t%MOD)%MOD;
dp2[k][j] = (dp2[k][j] + dp2[k][j-1] * t%MOD)%MOD;
}
dp2[k][1] = (dp2[k][1] + dp[i][k])%MOD;
dp[i][k] = (dp[i][k] + ans)%MOD;
}
FOR (j, 1, m) pref[i][j] = (pref[i][j-1] + dp[i][j])%MOD;
}
/*
3
1 5
5 10
5 15
*/
// FOR (i, 1, n){
// FOR (j, 1, m + 1) cout << dp[i][j] << ' ';
// cout << '\n';
// }
int ans = 0;
FOR (i, 1, n){
ans = (ans + dp[i][m+1])%MOD;
FOR (k, L[i], R[i]-1) ans = (ans + dp[i][k])%MOD;
}
cout << ans;
return 0;
}
