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 long long MX = (1LL << 60), MOD = 998244353;
long long n, INV;
long long bm;
vector<pair<long long, long long> > segments;
struct Seg {
  vector<long long> x[61], y[61];
  void add(long long l, long long r, long long i = 0, long long j = MX - 1,
           long long d = 60) {
    if (r < i || l > j) return;
    y[d].push_back(i);
    if (l <= i && r >= j) {
      x[d].push_back(i);
    } else {
      long long m = (i + j) / 2;
      add(l, r, i, m, d - 1);
      add(l, r, m + 1, j, d - 1);
    }
  }
} segA, segB;
long long modpow(long long x_, long long N_) {
  if (N_ == 0) return 1;
  long long a = modpow(x_, N_ / 2);
  a = (a * a) % MOD;
  if (N_ % 2) a = (a * x_) % MOD;
  return a;
}
long long inv(long long x_) { return modpow(x_, MOD - 2); }
long long calcSum(long long x) {
  return (x % MOD) * (x % MOD + 1) % MOD * INV % MOD;
}
long long calcSum(long long x, long long y) {
  return (calcSum(y) - calcSum(x) + MOD) % MOD;
}
long long addSeg(long long a, long long b) {
  long long c = ((a ^ b) | bm);
  long long d = (c ^ bm);
  segments.push_back({d, c});
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  INV = inv(2);
  for (int i = int(0); i < int(n); i++) {
    long long l, r;
    cin >> l >> r;
    segA.add(l, r);
  }
  cin >> n;
  for (int i = int(0); i < int(n); i++) {
    long long l, r;
    cin >> l >> r;
    segB.add(l, r);
  }
  for (int i = int(0); i < int(61); i++) {
    bm = ((1LL << i) - 1);
    ;
    for (long long a : segA.x[i])
      for (long long b : segB.x[i]) addSeg(a, b);
    for (long long a : segA.x[i])
      for (long long b : segB.y[i]) addSeg(a, b);
    for (long long a : segA.y[i])
      for (long long b : segB.x[i]) addSeg(a, b);
  }
  sort(segments.begin(), segments.end());
  long long mostR = 0, ans = 0;
  for (pair<long long, long long> p : segments) {
    mostR = max(mostR, p.first - 1);
    if (mostR > p.second) continue;
    ans += calcSum(mostR, p.second);
    mostR = p.second;
  }
  ans %= MOD;
  cout << ans << endl;
}
