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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
inline void wri(long long x) {
  write(x);
  putchar(' ');
}
const int maxn = 60 * 100 * 8 + 233;
const int mod = 998244353, inv2 = (mod + 1) / 2;
struct T {
  int rt, tot, c[maxn][2];
  bool tag[maxn];
  void insert(int &o, long long l, long long r, long long x, long long y) {
    if (!o) o = ++tot;
    if (x <= l && r <= y) {
      tag[o] = 1;
      return;
    }
    if (x < ((l + r) >> 1)) insert(c[o][0], l, ((l + r) >> 1), x, y);
    if (y > ((l + r) >> 1)) insert(c[o][1], ((l + r) >> 1), r, x, y);
  }
} A, B;
int nA, nB;
vector<pair<int, int> > q[61];
int bfs(int d, long long s) {
  for (auto x : q[d])
    if (A.tag[x.first] || B.tag[x.second])
      return (s + ((1ll << d) - 1) % mod * inv2) % mod * ((1ll << d) % mod) %
             mod;
  int res = 0;
  for (int i = (0); i < int(2); ++i) {
    q[d - 1].clear();
    for (auto x : q[d])
      for (int j = (0); j < int(2); ++j)
        if (A.c[x.first][j] && B.c[x.second][j ^ i])
          q[d - 1].push_back({A.c[x.first][j], B.c[x.second][i ^ j]});
    if (q[d - 1].size())
      res = (res + bfs(d - 1, s | ((long long)i << (d - 1)))) % mod;
  }
  return res;
}
int main() {
  nA = read();
  for (int i = (1); i <= int(nA); ++i) {
    long long l = read(), r = read() + 1;
    A.insert(A.rt, 0, 1ll << 60, l, r);
  }
  nB = read();
  for (int i = (1); i <= int(nB); ++i) {
    long long l = read(), r = read() + 1;
    B.insert(B.rt, 0, 1ll << 60, l, r);
  }
  q[60].push_back({A.rt, B.rt});
  writeln(bfs(60, 0));
  return 0;
}
