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题目大意

　　若$L_{i}\leqslant x_{i} \leqslant R_{i}$，求$\sum x_{i} = 0$以及$\sum x_{i} < 0$的方案数。$(L_{i}R_{i} \geqslant 0)$（好吧。是概率）

 /**

  * 51nod

  * Problem#1667

  * Accepted

  * Time: 187ms

  * Memory: 2052k

  */

 #include <bits/stdc++.h>

 using namespace std;

 typedef bool boolean;

 const int M = 1e9 + ;

 const signed int inf = (signed) (~0u >> );

 void exgcd(int a, int b, int& x, int& y) {

     if (!b)

         x = , y = ;

     else {

         exgcd(b, a % b, y, x);

         y -= (a / b) * x;

     }

 }

 int inv(int a, int n) {

     int x, y;

     exgcd(a, n, x, y);

     return (x < ) ? (x + n) : (x);

 }

 int add(int a, int b) {

     a += b;

     if (a < )

         a += M;

     if (a >= M)

         a -= M;

     return a;

 }

 int invs[];

 int n, m;

 int K = ;

 int rw = , req = , rl, rall = ;

 int ar[];

 inline void prepare() {

     for (int i = ; i <= ; i++)

         invs[i] = inv(i, M);

 }

 inline void init() {

     scanf("%d", &n);

     rw = req = , rall = , K = ;

     for (int i = , l, r; i < n; i++)

         scanf("%d%d", &l, &r), ar[i] = r - l, K -= l, rall = rall * 1ll * (ar[i] + ) % M;

     scanf("%d", &m);

     for (int i = , l, r; i < m; i++)

         scanf("%d%d", &l, &r), ar[n + i] = r - l, K += r, rall = rall * 1ll * (ar[n + i] + ) % M;

     n += m;

 }

 int C(int n, int m) {

     int rt = ;

     for (int i = ; i < m; i++)

         rt = rt * 1ll * (n - i) % M;

     for (int i = ; i <= m; i++)

         rt = rt * 1ll * invs[i] % M;

     return rt;

 }

 int calc(int dep, int sign, int K) {

     if (K < )

         return ;

     if (dep == -)

         return sign * C(K + n - , n - );

     return add(calc(dep - , -sign, K - ar[dep] - ), calc(dep - , sign, K));

 }

 inline void solve() {

     req = calc(n - , , K);

     ar[n++] = inf;

     rw = calc(n - , , K);

     rl = add(rall, -rw);

     rw = add(rw, -req);

 //    cerr << rl << " " << req << " " << rw << endl;

     rw = rw * 1ll * inv(rall, M) % M;

     req = req * 1ll * inv(rall, M) % M;

     rl = rl * 1ll * inv(rall, M) % M;

     printf("%d %d %d\n", rl, req, rw);

 }

 int T;

 int main() {

     scanf("%d", &T);

     prepare();

     while (T--) {

         init();

         solve();

     }

     return ;

 }