#include <bits/stdc++.h>
using namespace std;
//*/
#include <atcoder/all>
using namespace atcoder;
//*/
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#pragma GCC target("avx,avx2,fma")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#define rep(i,n) for(int i=0;i<n;i++)
#define Rep(i,a,b) for(int i=a;i<b;i++)
#define ALL(x) (x).begin(),(x).end()
#define dbgv(x); for(auto now : x) cout << now << " "; cout << endl;
//using P = pair<int,int>;
using ll = long long;
using ull = unsigned long long;
//*/
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<int> vint;
random_device rnd;
mt19937 rng(rnd());

using mint = modint998244353;
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}

void solve(){
  mint ans = 1;
  int al = 0;
  int k; cin >> k;
  vint l(k),m(k); rep(i,k) cin >> l[i] >> m[i];
  rep(i,k) al += l[i]*m[i];
  rep(i,k){
    int al2 = l[i]*m[i];
    ans *= comb(al,l[i]*m[i]);
    rep(j,m[i]){
      ans *= (mint)(comb(al2-1,l[i]-1));
      al2 -= l[i];
      if(al2==0) break;
      //cerr << al  << " " << l[i] << " " << m[i] << endl;
    }
    al -= l[i]*m[i];
  }
  //cout << al << emdl;
  cout << ans.val() << endl;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1; //cin >> t;
  rep(testcase,t) solve();
}