You are given two sets of integers: A and B. You need to output the sum of elements in the set C = \\{x | x = a ⊕ b, a ∈ A, b ∈ B\} modulo 998244353, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Each number should be counted only once.

For example, if A = \{2, 3\} and B = \{2, 3\} you should count integer 1 only once, despite the fact that you can get it as 3 ⊕ 2 and as 2 ⊕ 3. So the answer for this case is equal to 1 + 0 = 1.

Let's call a segment [l; r] a set of integers \\{l, l+1, ..., r\}.

The set A is given as a union of n_A segments, the set B is given as a union of n_B segments.

Input

The first line contains a single integer n_A (1 ≤ n_A ≤ 100).

The i-th of the next n_A lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^{18}), describing a segment of values of set A.

The next line contains a single integer n_B (1 ≤ n_B ≤ 100).

The i-th of the next n_B lines contains two integers l_j and r_j (1 ≤ l_j ≤ r_j ≤ 10^{18}), describing a segment of values of set B.

Note that segments in both sets may intersect.

Output

Print one integer — the sum of all elements in set C = \\{x | x = a ⊕ b, a ∈ A, b ∈ B\} modulo 998244353.

Examples

Input


2
3 5
5 8
3
1 2
1 9
2 9


Output


112


Input


1
1 9
2
2 4
2 10


Output


120

Note

In the second example, we can discover that the set C = \{0,1,...,15\}, which means that all numbers between 0 and 15 can be represented as a ⊕ b.
#include <bits/stdc++.h>
using namespace std;
int na, nb;
int rt[2], tg[1000006], ch[1000006][2], tot;
void ins(int &u, long long l, long long r, long long ql, long long qr) {
  if (l > qr || ql > r) return;
  if (!u) u = ++tot;
  if (ql <= l && r <= qr) return (void)(tg[u] = 1);
  long long m = l + r >> 1;
  ins(ch[u][0], l, m, ql, qr), ins(ch[u][1], m + 1, r, ql, qr);
}
int ans;
vector<pair<int, int> > q[66];
void dfs(long long offset, int d) {
  for (auto u : q[d])
    if (tg[u.first] || tg[u.second])
      return (void)(ans = (ans +
                           (offset + ((1ll << d) - 1) % 998244353 * 499122177) %
                               998244353 * ((1ll << d) % 998244353)) %
                          998244353);
  for (int x = (0), LIM = (1); x <= LIM; x++) {
    q[d - 1].clear();
    for (auto u : q[d])
      for (int y = (0), LIM = (1); y <= LIM; y++)
        if (ch[u.first][y] && ch[u.second][y ^ x])
          q[d - 1].push_back(make_pair(ch[u.first][y], ch[u.second][y ^ x]));
    if (q[d - 1].size()) dfs(offset | x * 1ll << (d - 1), d - 1);
  }
}
int main() {
  scanf("%d", &na);
  for (int i = (1), LIM = (na); i <= LIM; i++) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    ins(rt[0], 0, (1ll << 60) - 1, l, r);
  }
  scanf("%d", &nb);
  for (int i = (1), LIM = (nb); i <= LIM; i++) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    ins(rt[1], 0, (1ll << 60) - 1, l, r);
  }
  q[60].push_back(make_pair(rt[0], rt[1]));
  dfs(0, 60);
  printf("%d\n", (ans + 998244353) % 998244353);
}
