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>

using namespace std;
using ll = long long;

const int MOD = 1e9 + 7;
#define REP(i, n)  for(int i = 0; i < n; ++i)
#define FOR(i, t, n)  for(ll i = t; i <= (ll)n; ++i)
struct edge {
    int from, to;
    ll cost;
};


class Eratosthenes {
public:
    vector<bool> prime;

    Eratosthenes(int size) {
        prime.resize(size + 1, 1);
        prime[0] = 0;
        prime[1] = 0;
        FOR(i, 2, size) {
            if (prime[i]) {
                for (int j = 2 * i; j <= size; j += i) {
                    prime[j] = 0;
                }
            }
        }
    }

};


class Combination {
    vector<ll> fac; //n!(mod M)
    vector<ll> ifac; //k!^{M-2} (mod M)
public:
    Combination(ll size) {
        fac.resize(size);
        ifac.resize(size);
        fac[0] = 1;
        ifac[0] = 1;
        for (ll i = 0; i < size; i++) {
            fac[i + 1] = fac[i] * (i + 1) % MOD;
            ifac[i + 1] = ifac[i] * mpow(i + 1, MOD - 2) % MOD;
        }
    }

    ll mpow(ll x, ll n) {
        ll ans = 1;
        while (n != 0) {
            if (n & 1) ans = ans * x % MOD;
            x = x * x % MOD;
            n = n >> 1;
        }
        return ans;
    }

    ll comb(ll a, ll b) {
        if (a == 0 && b == 0)return 1;
        if (a < b || a < 0)return 0;
        ll tmp = ifac[a - b] * ifac[b] % MOD;
        return tmp * fac[a] % MOD;
    }
};


ll GCD(ll a, ll b) {
    if (b == 0)return a;
    return GCD(b, a % b);
}

ll N, K;
ll dp[1010101];


int main() {
    cin >> N;
    vector<int> A(N), T(N);
    REP(i, N)cin >> T[i];
    REP(i, N)cin >> A[i];
    if (T[N - 1] != A[0]) {
        cout << 0 << endl;
        return 0;
    }
    ll ans = 1;
    FOR(i, 1, N - 2) {
        if (T[i - 1] == T[i] && A[i] == A[i + 1]) {
            ans = ans * min(T[i], A[i]) % MOD;
        } else if (T[i - 1] == T[i] && A[i] > A[i + 1]) {
            if (T[i] < A[i]) {
                cout << 0 << endl;
                return 0;
            }
        } else if (T[i - 1] < T[i] && A[i] == A[i + 1]) {
            if (T[i] > A[i]) {
                cout << 0 << endl;
                return 0;
            }
        } else if (T[i - 1] < T[i] && A[i] > A[i + 1]) {
            if (T[i] != A[i]) {
                cout << 0 << endl;
                return 0;
            }
        }
    }
    cout << ans << endl;

    return 0;
}