#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MOD = 1e9 + 7;
int inv[505];
// Precompute modular inverses in O(N)
void precompute_inv(int n) {
inv[1] = 1;
for (int i = 2; i <= n; i++) {
inv[i] = MOD - (MOD / i) * inv[MOD % i] % MOD;
}
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector<int> a(n + 1), b(n + 1), coords;
for (int i = 1; i <= n; i++) {
cin >> a[i] >> b[i];
coords.push_back(a[i]);
coords.push_back(b[i] + 1);
}
// Coordinate Compression
sort(coords.begin(), coords.end());
coords.erase(unique(coords.begin(), coords.end()), coords.end());
precompute_inv(n + 1);
int m = coords.size();
// dp[i][j] = ways using a subset of first i schools,
// where school i MUST send boats in interval j.
vector<vector<int>> dp(n + 1, vector<int>(m, 0));
// pref[i][j] stores the sum of dp[0...i][j] for transitions
vector<vector<int>> pref(n + 1, vector<int>(m, 0));
// Base case: 1 way to send 0 boats (empty set)
for (int j = 0; j < m; j++) pref[0][j] = 1;
for (int j = 1; j < m; j++) { // For each interval [coords[j-1], coords[j])
int W = coords[j] - coords[j - 1];
for (int i = 1; i <= n; i++) {
// Only process if school i can actually fit in this interval
if (a[i] <= coords[j - 1] && b[i] >= coords[j] - 1) {
int c = 1;
// Ways to pick 1 school from interval W: C(W, 1)
int current_comb = W;
for (int p = i - 1; p >= 0; p--) {
// Transition:
// (Ways from previous schools in smaller intervals) * (Ways to clump schools in this interval)
int ways = (pref[p][j - 1] * current_comb) % MOD;
dp[i][j] = (dp[i][j] + ways) % MOD;
// If the previous school p could also fit in this interval,
// we increment c and update our combination: C(W+c-1, c)
if (p > 0 && a[p] <= coords[j - 1] && b[p] >= coords[j] - 1) {
c++;
current_comb = current_comb * (W + c - 1) % MOD;
current_comb = current_comb * inv[c] % MOD;
}
}
}
}
// After processing interval j for all schools, update prefix sums for the next interval
for (int i = 1; i <= n; i++) {
pref[i][j] = (pref[i - 1][j] + pref[i][j - 1] - pref[i - 1][j - 1] + dp[i][j] + MOD) % MOD;
// Simplified prefix update:
// This just needs to track sum of all dp[0...i][0...j]
}
// Let's use a cleaner prefix sum for "all ways using schools 0...i in intervals < j"
int total_prev_intervals = 0;
for (int i = 0; i <= n; i++) {
// sum of dp[i][0...j]
pref[i][j] = (pref[i][j-1] + dp[i][j]) % MOD;
}
}
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j < m; j++) {
ans = (ans + dp[i][j]) % MOD;
}
}
cout << ans << endl;
return 0;
}
// WTF
