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 = 1e6 + 10;
const int mod = 998244353;
const int inv2 = (mod + 1) >> 1;
struct SegTree {
  int ch[N][2], tot = 0;
  bool tag[N];
  void ins(int &i, long long l, long long r, long long L = 0,
           long long R = 1ll << 60) {
    if (!i) i = ++tot;
    if (l == L && r == R) {
      tag[i] = 1;
      return;
    }
    long long mid = L + R >> 1;
    if (l < mid) ins(ch[i][0], l, r < mid ? r : mid, L, mid);
    if (r > mid) ins(ch[i][1], l > mid ? l : mid, r, mid, R);
  }
} A, B;
std::vector<pair<int, int> > q[70];
long long dfs(int d, long long offset) {
  for (auto x : q[d])
    if (A.tag[x.first] || B.tag[x.second])
      return (offset + ((1ll << d) - 1) % mod * inv2) % mod *
             ((1ll << d) % mod) % mod;
  long long res = 0;
  for (int x = 0; x < 2; x++) {
    q[d - 1].clear();
    for (auto z : q[d])
      for (int y = 0; y < 2; y++)
        if (A.ch[z.first][y] && B.ch[z.second][x ^ y])
          q[d - 1].push_back({A.ch[z.first][y], B.ch[z.second][x ^ y]});
    if (q[d - 1].size())
      res = (res + dfs(d - 1, offset | ((long long)x << d - 1))) % mod;
  }
  return res;
}
int rtA, rtB;
int main(int argc, char const *argv[]) {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long l, r;
    scanf("%I64d%I64d", &l, &r);
    A.ins(rtA, l, r + 1);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    long long l, r;
    scanf("%I64d%I64d", &l, &r);
    B.ins(rtB, l, r + 1);
  }
  q[60].push_back({rtA, rtB});
  printf("%I64d\n", dfs(60, 0));
  return 0;
}
