#include <bits/stdc++.h>
using namespace std;

void gi(int &x) {char ch = getchar(); x = 0; while (ch < '0' || ch > '9') ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();}
void pi(int x) {if (x > 9) pi(x / 10); putchar(x % 10 + 48);}

//begin persistent segment tree
const int sz = 5000000;
int a[sz], l[sz], r[sz], N, tn;
int add(int x, int p, int s = tn) {
  int u = ++N; a[u] = a[x] + 1; l[u] = l[x]; r[u] = r[x];
  if (s == 1) return u;
  else if (p < (s >> 1)) l[u] = add(l[x], p, s >> 1);
  else r[u] = add(r[x], p - (s >> 1), s >> 1);
  return u;
}
int sum(int x, int p, int s = tn) {
  if (s == 1) return a[x];
  else if (p < (s >> 1)) return sum(l[x], p, s >> 1);
  else return a[l[x]] + sum(r[x], p - (s >> 1), s >> 1);
}
//end persistent segment tree

int n, q, cnt, x[233333], s[233333], root[233333];//x is query; s is used in brute force; root is the roots of persistent segment tree
pair <int, int> itv[233333];//intervals
#define L first
#define R second
int B;

//actually query (l1+1, r1, l2+1, r2)
int query(int l1, int r1, int l2, int r2) {
  return sum(root[r1], r2) + sum(root[l1], l2) - sum(root[l1], r2) - sum(root[r1], l2);
}

void doit() {
  int i, j, ans;
  N = 0; gi(n); for (tn = 1; tn <= n; tn <<= 1);
  for (i = 1; i <= n; i++) gi(itv[i].L), gi(itv[i].R);
  sort(itv + 1, itv + n + 1);
  for (i = j = 1; i <= n; i++) {
    root[i] = root[i - 1];
    for (; j <= n && itv[j].L <= i; j++)
      root[i] = add(root[i], itv[j].R);
  }
  B = int(sqrt(n / log(n)) * 0.8);
  
  gi(q);
  while (q--) {
    gi(cnt); ans = 0;
    for (i = 1; i <= cnt; i++) gi(x[i]);
    if (cnt <= B) {
      sort(x + 1, x + cnt + 1); x[cnt + 1] = n + 1;
      for (i = 1; i <= cnt; i++)
	for (j = i; j <= cnt; j += 2)
	  ans += query(x[i - 1], x[i], x[j] - 1, x[j + 1] - 1);
    } else {
      for (i = 1; i <= n; i++) s[i] = 0;
      for (i = 1; i <= cnt; i++) s[x[i]]++;
      for (i = 1; i <= n; i++) s[i] += s[i - 1];
      for (i = 1; i <= n; i++) ans += (1 & (s[itv[i].R] ^ s[itv[i].L - 1]));
    }
    pi(ans); putchar('\n');
  }
}

int main() {int T; gi(T); while (T--) doit(); return 0;}