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;
int n, m, rot0, rot1;
struct SGT {
  int ch[1000005][2], tot = 0;
  bool tag[1000005];
  void Insert(int &q, long long l, long long r, long long L = 0,
              long long R = 1ll << 60) {
    if (!q) q = ++tot;
    if (l <= L && r >= R) {
      tag[q] = 1;
      return;
    }
    long long mid = (L + R) >> 1;
    if (l < mid) Insert(ch[q][0], l, r, L, mid);
    if (r > mid) Insert(ch[q][1], l, r, mid, R);
  }
} A, B;
vector<pair<int, int> > q[70];
long long Getans(int d, long long s) {
  for (int i = 0; i < (int)q[d].size(); i++) {
    pair<int, int> x = q[d][i];
    if (A.tag[x.first] || B.tag[x.second])
      return (s + ((1ll << d) - 1) % 998244353 * 499122177) % 998244353 *
             ((1ll << d) % 998244353) % 998244353;
  }
  long long res = 0;
  for (int x = 0; x < 2; x++) {
    q[d - 1].clear();
    for (int i = 0; i < (int)q[d].size(); i++) {
      pair<int, int> z = q[d][i];
      for (int y = 0; y < 2; y++)
        if (A.ch[z.first][y] && B.ch[z.second][x ^ y])
          q[d - 1].push_back(
              make_pair(A.ch[z.first][y], B.ch[z.second][x ^ y]));
    }
    if (q[d - 1].size())
      res = (res + Getans(d - 1, s | ((long long)x << (d - 1)))) % 998244353;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    A.Insert(rot0, l, r + 1);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    long long l, r;
    scanf("%lld%lld", &l, &r);
    B.Insert(rot1, l, r + 1);
  }
  q[60].push_back(make_pair(rot0, rot1));
  printf("%lld\n", Getans(60, 0));
}
