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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit(b) (1ll << (b))
#define uniq(v) (v).erase(unique(all(v)), (v).end())
#define rsort(v) sort(all(v)); reverse(all(v))

using namespace std;

using i32 = int;
using i64 = long long;
using f80 = long double;
using vi32 = vector<i32>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << '\n'; exit(0); }
inline void no() { cout << "No" << '\n'; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }
template<typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template<typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template<typename T> inline void amax(T &x, T y) { if (x < y) x = y; }
template<typename T> inline void amin(T &x, T y) { if (x > y) x = y; }
template<typename T> inline T power(T x, i64 n) { T r = 1; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template<typename T> istream& operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }

template<int mod>
struct ModInt {
  int x;
  ModInt(): x(0) {}
  ModInt(i64 a) { x = a % mod; if (x < 0) x += mod; }
  ModInt &operator+=(ModInt that) { x = (x + that.x) % mod; return *this; }
  ModInt &operator-=(ModInt that) { x = (x + mod - that.x) % mod; return *this; }
  ModInt &operator*=(ModInt that) { x = (i64) x * that.x % mod; return *this; }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt operator-() { return ModInt(-this->x); }
  friend ostream& operator<<(ostream &os, ModInt m) { return os << m.x; }
  ModInt inverse() {
    int a = x, b = mod, u = 1, v = 0;
    while (b) { int t = a / b; a -= t * b; u -= t * v; swap(a, b); swap(u, v); }
    return ModInt(u);
  }
  #define op(o, p) ModInt operator o(ModInt that) { return ModInt(*this) p that; }
    op(+, +=) op(-, -=) op(*, *=) op(/, /=)
  #undef op
  #define op(o) bool operator o(ModInt that) const { return x o that.x; }
    op(==) op(!=) op(<) op(<=) op(>) op(>=)
  #undef op
};

using mint = ModInt<1000000007>;

void solve() {
  int N; cin >> N;
  vi32 T(N), A(N); cin >> T >> A;
  mint ans = 1;
  rep(i, N) {
    bool P = false, Q = false;
    if (i == 0 || i > 0 && T[i - 1] < T[i]) P = true;
    if (i == N - 1 || i < N - 1 && A[i] > A[i + 1]) Q = true;
    if (P) {
      if (Q) {
        if (T[i] != A[i]) ans = 0;
      } else {
        if (T[i] > A[i]) ans = 0;
      }
    } else {
      if (Q) {
        if (T[i] < A[i]) ans = 0;
      } else {
        ans *= min(T[i], A[i]);
      }
    }
  }
  cout << ans << '\n';
}