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 mod = 998244353, inv2 = (mod + 1) / 2;
int n, m, rt[2], ch[60 * 100 * 8][2], cnt, ans;
bool Real[60 * 100 * 8];
void ins(int &i, long long l, long long r, long long x, long long y) {
  if (!i) i = ++cnt;
  if (x <= l && r <= y) return void(Real[i] = 1);
  long long mid = (l + r) >> 1;
  if (x <= mid) ins(ch[i][0], l, mid, x, y);
  if (y > mid) ins(ch[i][1], mid + 1, r, x, y);
}
vector<pair<int, int> > q[65];
void solve(long long x, int d) {
  for (pair<int, int> now : q[d])
    if (Real[now.first] || Real[now.second])
      return void(ans = (ans + (x % mod + ((1ll << d) % mod - 1) * inv2 % mod) *
                                   ((1ll << d) % mod)) %
                        mod);
  for (int i = 0; i < 2; i++) {
    q[d - 1].clear();
    for (pair<int, int> now : q[d])
      for (int j = 0; j < 2; j++)
        if (ch[now.first][j] && ch[now.second][i ^ j])
          q[d - 1].push_back(
              pair<int, int>(ch[now.first][j], ch[now.second][i ^ j]));
    if (q[d - 1].size()) solve(x | (i ? 1ll << (d - 1) : 0), d - 1);
  }
}
int main() {
  scanf("%d", &n);
  long long x, y;
  while (n--) scanf("%lld%lld", &x, &y), ins(rt[0], 0, (1ll << 60) - 1, x, y);
  scanf("%d", &m);
  while (m--) scanf("%lld%lld", &x, &y), ins(rt[1], 0, (1ll << 60) - 1, x, y);
  q[60].push_back(pair<int, int>(rt[0], rt[1]));
  solve(0, 60);
  printf("%d\n", ans);
}
