Problem:
Score : 400 points

Problem Statement

Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range.
The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N.
Mr. Takahashi traversed the range from the west and Mr. Aoki from the east.

The height of Mt. i is h_i, but they have forgotten the value of each h_i.
Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i).
Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i.

We know that the height of each mountain h_i is a positive integer.
Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7.

Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights.
In such a case, output 0.

Constraints

1 ≤ N ≤ 10^5

1 ≤ T_i ≤ 10^9

1 ≤ A_i ≤ 10^9

T_i ≤ T_{i+1} (1 ≤ i ≤ N - 1)

A_i ≥ A_{i+1} (1 ≤ i ≤ N - 1)

Input

The input is given from Standard Input in the following format:

N
T_1 T_2 ... T_N
A_1 A_2 ... A_N

Output

Print the number of possible sequences of the mountains' heights, modulo 10^9 + 7.

Sample Input 1

5
1 3 3 3 3
3 3 2 2 2

Sample Output 1

4

The possible sequences of the mountains' heights are:

1, 3, 2, 2, 2

1, 3, 2, 1, 2

1, 3, 1, 2, 2

1, 3, 1, 1, 2

for a total of four sequences.

Sample Input 2

5
1 1 1 2 2
3 2 1 1 1

Sample Output 2

0

The records are contradictory, since Mr. Takahashi recorded 2 as the highest peak after climbing all the mountains but Mr. Aoki recorded 3.

Sample Input 3

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5

Sample Output 3

884111967

Don't forget to compute the number modulo 10^9 + 7.

Sample Input 4

1
17
17

Sample Output 4

1

Some mountain ranges consist of only one mountain.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

// boost
#include <boost/multiprecision/cpp_int.hpp>
using boost::multiprecision::cpp_int;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;
using vld = vector<ld>;
using vvld = vector<vld>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

#define all(v) v.begin(), v.end()

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

template <class T>
inline void chmax(T& a, T b) {
  a = std::max(a, b);
}
template <class T>
inline void chmin(T& a, T b) {
  a = std::min(a, b);
}

#define mod (ll)(1e9 + 7)
// #define mod (998244353ll)

const long long INF = 1LL << 60;

// #define mod (ll)(1e9 + 7)

template <ll ModVal>
struct ModInt {
  ll x;

  ModInt(ll _x = 0) : x((_x % ModVal + ModVal) % ModVal) {
  }

  ModInt operator-() const {
    return ModInt(-x);
  }
  ModInt& operator+=(const ModInt a) {
    x += a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator-=(const ModInt a) {
    x = x + ModVal - a.x;
    if (x >= ModVal)
      x -= ModVal;
    return *this;
  }
  ModInt& operator*=(const ModInt a) {
    x *= a.x;
    x %= ModVal;
    return *this;
  }

  // 返り値: gcd(a, b)
  // ax + by = gcd(a, b) を満たす (x, y)
  ll ext_gcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    ll tmp = a / b;
    ll d = ext_gcd(b, a - b * tmp, y, x);
    y -= tmp * x;
    return d;
  }

  // 逆元
  ModInt inv() {
    ll u, v;
    ext_gcd(x, ModVal, u, v);
    return ModInt(u);
  }

  ModInt inv(const ModInt a) {
    ll u, v;
    ext_gcd(a.x, ModVal, u, v);
    return ModInt(u);
  }

  ModInt& operator/=(const ModInt a) {
    return (*this) *= inv(a);
  }

  ModInt operator+(const ModInt a) const {
    ModInt retval(*this);
    return retval += a;
  }
  ModInt operator-(const ModInt a) const {
    ModInt retval(*this);
    return retval -= a;
  }
  ModInt operator*(const ModInt a) const {
    ModInt retval(*this);
    return retval *= a;
  }
  ModInt operator/(const ModInt a) const {
    ModInt retval(*this);
    return retval /= a;
  }

  ModInt pow(ll n) {
    ModInt ans(1);
    while (n) {
      if (n & 1)
        ans = ans * x;
      *this = (*this) * (*this);
      n = n >> 1;
    }
    return ans;
  }

  constexpr const ll& value() {
    return this->x;
  }
};

template <ll ModVal>
ostream& operator<<(ostream& os, const ModInt<ModVal>& a) {
  os << a.x;
  return os;
}

using mint = ModInt<mod>;

template <typename T>
class Combination {
 public:
  Combination(ll _max_n) : max_n(_max_n), factional(max_n + 1), inv(max_n + 1) {
    factional[0] = 1;
    inv[0] = 1;
    for (ll i = 0; i < max_n; i++) {
      factional[i + 1] = factional[i] * (i + 1); // n!(mod M)
      inv[i + 1] = inv[i] / (i + 1);             // k!^(M-2) (mod M)
    }
  }

  // nCk
  T choose(ll n, ll k) {
    if (n == 0 && k == 0)
      return 1;
    if (n < k || n < 0)
      return 0;
    T tmp = inv[n - k] * inv[k];
    return tmp * factional[n];
  }

  T permutation(ll n, ll k) {
    if (n - k < 0) {
      return 0;
    }
    return factional[n] / factional[n - k];
  }

 private:
  const ll max_n;
  std::vector<T> factional;
  std::vector<T> inv;
};

using Comb = Combination<mint>;

using vm = vector<mint>;
using vvm = vector<vm>;
using vvvm = vector<vvm>;

mint solve() {
  ll n;
  cin >> n;
  vll t(n);
  vll a(n);
  rep(i, n) cin >> t[i];
  rep(i, n) cin >> a[i];

  vll h(n);
  vll mn(n, INF);
  vll mx(n, 1);

  h[n - 1] = a[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    if (a[i] > a[i + 1]) {
      h[i] = a[i];
    } else {
      mx[i] = a[i];
    }
  }

  if (h[0] && h[0] != t[0]) {
    return 0;
  }

  mint ans = 1;

  REP(i, n) {
    if (t[i] > t[i - 1]) {
      if (h[i]) {
        if (h[i] != t[i])
          return 0;
      } else {
        if (t[i] > mx[i]) {
          return 0;
        }
      }
    } else {
      if (h[i]) {
        if (t[i] < h[i]) {
          return 0;
        }
      } else {
        ans *= min(a[i], t[i]);
      }
    }
  }

  return ans;
}

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  cout << solve() << endl;

  return 0;
}