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;
const long long maxlog = 61, P = 998244353, inv2 = (P + 1) / 2;
vector<pair<long long, long long> > V[4][maxlog + 1], res;
void adda(long long x, long long L, long long R, long long l, long long r,
          long long dep) {
  if (L <= l && r <= R) {
    V[x][dep].push_back(make_pair(l, r));
    return;
  }
  V[x ^ 1][dep].push_back(make_pair(l, r));
  long long m = (l + r) >> 1;
  if (L <= m) adda(x, L, R, l, m, dep + 1);
  if (R > m) adda(x, L, R, m + 1, r, dep + 1);
}
signed main() {
  long long na, nb;
  cin >> na;
  long long l, r;
  for (long long i = 0; i < na; ++i) {
    cin >> l >> r;
    adda(0, l, r, 0, (1ll << maxlog) - 1, 0);
  }
  cin >> nb;
  for (long long i = 0; i < nb; ++i) {
    cin >> l >> r;
    adda(2, l, r, 0, (1ll << maxlog) - 1, 0);
  }
  for (long long d = 0; d <= maxlog; ++d) {
    for (auto pa : V[0][d])
      for (auto pb : V[2][d]) {
        res.push_back(make_pair(pa.first ^ pb.first, 1));
        res.push_back(make_pair((pa.first ^ pb.second) + 1, -1));
      }
    for (auto pa : V[0][d])
      for (auto pb : V[3][d]) {
        res.push_back(make_pair(pa.first ^ pb.first, 1));
        res.push_back(make_pair((pa.first ^ pb.second) + 1, -1));
      }
    for (auto pa : V[1][d])
      for (auto pb : V[2][d]) {
        res.push_back(make_pair(pa.first ^ pb.first, 1));
        res.push_back(make_pair((pa.first ^ pb.second) + 1, -1));
      }
  }
  sort(res.begin(), res.end());
  long long last = 0, cur = 0;
  long long ans = 0;
  for (auto p : res) {
    if (cur > 0) {
      (ans += ((p.first - last) % P) * ((p.first + last - 1) % P) % P * inv2) %=
          P;
    }
    last = p.first;
    cur += p.second;
  }
  cout << ans << endl;
}
