#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long f_pow(long long a, long long b) {
  long long res = 1, temp = a % mod;
  while (b) {
    if (b & 1) res = res * temp % mod;
    temp = temp * temp % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  int n;
  scanf("%d", &n);
  pair<long long, long long> p[200005];
  long long pre[200005];
  pre[0] = 1;
  p[0] = make_pair(0, 0);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &p[i].first);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &p[i].second);
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = pre[i - 1] * (p[i].second - p[i].first + 1) % mod;
  }
  long long lans1 = 1;
  long long ans1 = (p[1].second - p[1].first + 1);
  long long ans2 = (p[1].second - p[1].first + 1);
  for (int i = 2; i <= n; i++) {
    long long tot = (p[i].second - p[i].first + 1);
    long long l = max(p[i].first, p[i - 1].first);
    long long r = min(p[i].second, p[i - 1].second);
    long long sum = max(r - l + 1, 0ll);
    long long ll = max(max(p[i].first, p[i - 1].first), p[i - 2].first);
    long long rr = min(min(p[i].second, p[i - 1].second), p[i - 2].second);
    long long sum2 = max(rr - ll + 1, 0ll);
    if (i == 2)
      ans2 = (ans2 + 2 * ans1 + pre[i - 1]) * tot -
             2 * sum * (lans1)-sum * pre[i - 2];
    else
      ans2 = (ans2 + 2 * ans1 + pre[i - 1]) * tot -
             2 * (sum * (lans1 + pre[i - 2]) - sum2 * pre[i - 3]) -
             sum * pre[i - 2];
    ans2 %= mod;
    ans2 += mod;
    ans2 %= mod;
    lans1 = ans1;
    ans1 = ans1 * tot + pre[i] - pre[i - 2] * sum;
    ans1 %= mod;
  }
  printf("%lld\n", ans2 * f_pow(pre[n], mod - 2) % mod);
}
