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 int N = 63;
const long long MAX = (1ll << 60), MOD = 998244353, INV_2 = 499122177;
int n, m;
vector<pair<long long, long long> > tot[2][N], sub[2][N], ans;
void dfs(long long l, long long r, long long L, long long R, int depth,
         int typ) {
  tot[typ][depth].push_back(pair<long long, long long>(l, r));
  if (L <= l && r <= R) {
    sub[typ][depth].push_back(pair<long long, long long>(l, r));
    return;
  }
  long long mid = (l + r) >> 1ll;
  if (L <= mid) dfs(l, mid, L, R, depth + 1, typ);
  if (R > mid) dfs(mid + 1, r, L, R, depth + 1, typ);
}
void modify(long long &sum, long long l, long long r) {
  long long t_sum = (l + r) % MOD * ((r - l + 1) % MOD) % MOD * INV_2 % MOD;
  sum += t_sum;
  if (sum >= MOD) sum -= MOD;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    dfs(0, MAX - 1, l, r, 1, 0);
  }
  scanf("%d", &m);
  for (int i = (1); i <= (m); ++i) {
    long long l, r;
    scanf("%lld %lld", &l, &r);
    dfs(0, MAX - 1, l, r, 1, 1);
  }
  ans.clear();
  for (int typ = (0); typ <= (1); ++typ)
    for (int i = (1); i <= (61); ++i)
      for (auto x : tot[typ][i])
        for (auto y : sub[!typ][i])
          ans.push_back(pair<long long, long long>(
              x.first ^ y.first, (x.first ^ y.first) + (x.second - x.first)));
  sort(ans.begin(), ans.end());
  long long sum = 0, cur = 0;
  for (auto x : ans) {
    cur = max(cur, x.first - 1);
    if (x.second > cur) modify(sum, cur + 1, x.second), cur = x.second;
  }
  printf("%lld\n", sum);
  return 0;
}
