#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int N;
int L[202020],R[202020];
const ll mo=998244353;

ll modpow(ll a, ll n = mo-2) {
	ll r=1;a%=mo;
	while(n) r=r*((n%2)?a:1)%mo,a=a*a%mo,n>>=1;
	return r;
}

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	ll rank=0;
	FOR(i,N) {
		cin>>L[i]>>R[i];
		if(i) {
			ll TL=min(max(L[0],L[i]),R[0]+1);
			ll TR=min(R[0],R[i]);
			if(TL>L[0]) (rank+=TL-L[0])%=mo;
			ll a=(R[i]-TL+R[i]-TR)*(TR-TL+1);
			if(max(L[i],L[0])<=min(R[0],R[i])) (rank+=a%mo*modpow(2*(R[i]-L[i]+1)))%=mo;
			
		}
	}
	rank = (rank*modpow(R[0]-L[0]+1)+1)%mo;
	cout<<rank<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}