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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        n: usize,
        t: [i64; n],
        a: [i64; n],
    };
    let mut f = vec![0_i64; n];
    f[0] = t[0];
    let mut p = t[0];
    for (i, &t_i) in t.iter().enumerate().skip(1) {
        if p < t_i {
            f[i] = t_i;
        }
        p = t_i;
    }
    if f[n - 1] > 0 && f[n - 1] != a[n - 1] {
        println!("0");
        return;
    }
    f[n - 1] = a[n - 1];
    p = a[n - 1];
    for (i, &a_i) in a.iter().enumerate().rev().skip(1) {
        if p < a_i {
            if f[i] > 0 && f[i] != a_i {
                println!("0");
                return;
            }
            f[i] = a_i;
        }
        p = a_i;
    }

    let mut ans = 1_i64;
    for (&f_i, (&t_i, &a_i)) in f.iter().zip(t.iter().zip(a.iter())) {
        if f_i == 0 {
            ans *= std::cmp::min(t_i, a_i);
            ans %= 1_000_000_007;
        }
    }
    println!("{}", ans);
}