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: Accepted)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: usize,
        t: [i64; n],
        a: [i64; n]
    }
    const M: i64 = 1000000007;

    let mut th = Vec::new();
    let mut tmax = t[0];
    th.push((tmax, tmax));
    for i in 1..n {
        if t[i] > tmax {
            tmax = t[i];
            th.push((tmax, tmax));
        } else {
            th.push((1, tmax));
        }
    }
    let mut ah = Vec::new();
    let mut amax = a[n-1];
    ah.push((amax, amax));
    for i in (0..n-1).rev() {
        if a[i] > amax {
            amax = a[i];
            ah.push((amax, amax));
        } else {
            ah.push((1, amax));
        }
    }
    ah.reverse();

    let mut ans = 1i64;
    for i in 0..n {
        if let Some((h1, h2)) = inner(th[i], ah[i]) {
            ans *= h2 + 1 - h1;
            ans %= M;
        } else {
            println!("0");
            return;
        }
    }
    println!("{}", ans);

}

fn inner(r1: (i64, i64), r2: (i64, i64)) -> Option<(i64, i64)> {
    if r1.0 <= r2.0 && r2.1 <= r1.1 {
        Some(r2)
    } else if r2.0 <= r1.0 && r1.1 <= r2.1 {
        Some(r1)
    } else {
        None
    }
}