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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
using vint = vector<int>;
using vvint = vector<vint>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vchar = vector<char>;
using vvchar = vector<vchar>;
using vp = vector<P>;
using vpp = vector<pair<P, P>>;
using vvp = vector<vp>;

#define rep(i, n) for (int i = 0; i < n; ++i)
#pragma region Debug
istream &operator>>(istream &is, P &a)
{
    return is >> a.first >> a.second;
}
ostream &operator<<(ostream &os, const P &a) { return os << "(" << a.first << "," << a.second << ")"; }

template <typename T>
void view(const std::vector<T> &v)
{
    for (const auto &e : v)
    {
        std::cout << e << " ";
    }
    std::cout << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>> &vv)
{
    for (const auto &v : vv)
    {
        view(v);
    }
}
#pragma endregion

#pragma region mod
const int p = 1e9 + 7; //10^9 + 7
struct mint
{
    ll x;
    mint(ll xx = 0) : x(xx % p) {}
    mint &operator+=(const mint a)
    {
        if ((x += a.x) >= p)
            x -= p;
        return *this;
    }
    mint &operator-=(const mint a)
    {
        if ((x += p - a.x) >= p)
            x -= p;
        return *this;
    }
    mint &operator*=(const mint a)
    {
        (x *= a.x) %= p;
        return *this;
    }

    mint operator+(const mint a) const
    {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const
    {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const
    {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const
    {
        mint r = 1;
        mint a = *this;
        while (true)
        {
            if (t & 1)
            {
                r *= a;
            }
            t >>= 1;
            if (t == 0)
                break;
            a *= a;
        }

        return r;
    }

    mint inv() const
    {
        return pow(p - 2);
    }

    mint &operator/=(const mint a)
    {
        return (*this) *= a.inv();
    }
};
istream &operator>>(istream &is, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

struct comb
{
    vector<mint> fact, invs;
    comb(int n) : fact(n + 1), invs(n + 1)
    {
        assert(n < p);

        fact[0] = 1;
        for (int i = 1; i <= n; ++i)
            fact[i] = fact[i - 1] * i;

        invs[n] = fact[n].inv();
        for (int i = n; i >= 1; --i)
            invs[i - 1] = invs[i] * i;
    }
    mint operator()(int n, int k)
    {
        if (k < 0 || k > n)
            return 0;
        return fact[n] * invs[n - k] * invs[k];
    }
};
#pragma endregion

int main()
{
    int n;
    cin >> n;
    vint t(n), a(n);
    rep(i, n) cin >> t[i];
    rep(i, n) cin >> a[i];
    int maxt = 0, maxa = 0;
    rep(i, n)
    {
        maxt = max(maxt, t[i]);
        maxa = max(maxa, a[i]);
    }
    if (maxt != maxa)
    {
        cout << 0 << endl;
        return 0;
    }
    if (t[0] > a[0] || a[n - 1] > t[n - 1])
    {
        cout << 0 << endl;
        return 0;
    }

    for (int i = 1; i < n - 1; i++)
    {
        if (t[i] > t[i - 1] && a[i] > a[i + 1] && a[i] != t[i])
        {
            cout << 0 << endl;
            return 0;
        }
    }
    reverse(a.begin(), a.end());
    for (int i = 1; i < n; i++)
    {
        if (t[i] < t[i - 1] || a[i] < a[i - 1])
        {
            cout << 0 << endl;
            return 0;
        }
    }
    reverse(a.begin(), a.end());

    mint ans = 1;
    for (int i = 1; i < n - 1; i++)
    {
        if (t[i] > t[i - 1])
            continue;
        if (a[i] > a[i + 1])
            continue;
        int x = min(t[i], a[i]);
        ans *= x;
    }

    cout << ans << endl;

    return 0;
}