/***********************************************
* auth: tapilyoca *
* date: 06/29/2025 at 19:58:10 *
* dots: https://github.com/tapilyoca/dotilyoca *
***********************************************/
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9+7;
template<typename T>
using vec = vector<T>;
using ll = long long;
using vll = vec<ll>;
using vvll = vec<vll>;
using pll = pair<ll,ll>;
using str = string;
#define pb push_back
#define dbg(x) if(1) cerr << #x << ": " << x << endl;
/***********************************************/
void solve() {
ll n;
cin >> n;
vec<pll> boatRanges(n);
ll mxy = -1;
for(int i = 0; i < n; i++) {
ll x, y;
cin >> x >> y;
boatRanges[i] = {x,y};
mxy = max(mxy,y);
}
vec<vec<ll>> dp(n,vec<ll>(mxy+1,0)); // dp[i][p] = ans up to i assuming person i sends out p
for(int p = boatRanges[0].first; p <= boatRanges[0].second; p++) {
dp[0][p] = 1;
}
dp[0][0] = 1;
for(int i = 1; i < n; i++) {
for(int p = boatRanges[i].first; p <= boatRanges[i].second; p++) {
ll ways = 0;
for(int j = 0; j < p; j++) {
ways += dp[i-1][j];
ways %= MOD;
}
dp[i][p] = ways;
}
// special case: dont send. answer is just sum of previous row
ll sum = 0;
for(int j = 0; j <= mxy; j++) {
sum += dp[i-1][j];
sum %= MOD;
}
dp[i][0] = sum;
}
ll ans = 0;
for(ll &x : dp[n-1]) {
ans += x;
ans %= MOD;
}
// cerr << "DP: " << endl;
// for(vll &x : dp){
// for(ll &y : x) {
// cerr << y << " ";
// } cerr << endl;
// }
cout << ans - 1 << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
while(t--) {
solve();
}
return 0;
}
