#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} using mint = modint998244353; int main(){ int n; cin >> n; vector l(n); vector r(n); rep(i, n) cin >> l[i] >> r[i]; vector d(n); rep(i, n) d[i] = r[i] - l[i] + 1; auto f = [&](int i, int x) -> mint{ if(x < l[i]) return mint(0); else if(x < r[i]){ mint width = x - l[i] + 1; return width * (width + 1) / 2 / d[i]; } { mint width = x - r[i]; return mint(d[i] + 1) / 2 + width; } }; mint ans = 1; for(int i = 1; i < n; i++){ mint prob = 0; prob += f(i, r[0]); prob -= f(i, l[0] - 1); ans += mint(1) - prob / d[0]; // cout << prob << "\n"; } cout << ans << "\n"; return 0; }