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;
long long L, R;
struct Trie {
  int son[100 * 500][2], tot = 1;
  bool tag[100 * 500];
  void insert(int &x, long long l = 0, long long r = (1ll << 60) - 1) {
    if (!x) x = ++tot;
    if (L <= l && R >= r) {
      tag[x] = 1;
      return;
    }
    long long mid = (l + r) >> 1;
    if (L <= mid) insert(son[x][0], l, mid);
    if (R > mid) insert(son[x][1], mid + 1, r);
  }
} A, B;
int rt = 1;
const long long inv2 = (998244353 + 1) >> 1;
struct Node {
  int a, b;
  Node(int x, int y) { a = x, b = y; }
};
vector<Node> q[61];
long long dfs(int d, long long pre) {
  for (auto x : q[d]) {
    if (A.tag[x.a] || B.tag[x.b])
      return ((pre << 1) + (1ll << d) - 1) % 998244353 *
             ((1ll << d) % 998244353) % 998244353 * inv2 % 998244353;
  }
  long long res = 0;
  for (int x = 0; x < 2; x++) {
    q[d - 1].clear();
    for (int y = 0; y < 2; y++) {
      for (auto v : q[d])
        if (A.son[v.a][y] && B.son[v.b][x ^ y])
          q[d - 1].emplace_back(A.son[v.a][y], B.son[v.b][x ^ y]);
    }
    if (!q[d - 1].empty())
      (res += dfs(d - 1, pre | (x * 1ll << (d - 1)))) %= 998244353;
  }
  return res;
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lld%lld", &L, &R), A.insert(rt);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) scanf("%lld%lld", &L, &R), B.insert(rt);
  q[60].emplace_back(1, 1);
  printf("%lld\n", dfs(60, 0));
  return 0;
}
