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(non_snake_case,unused_imports,dead_code)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;
use std::cmp::Ordering;

use proconio::input;
use proconio::marker::{Chars,Usize1,Isize1};

fn comp_f64(a: &f64, b: &f64) -> Ordering {
    if a < b {
        return Ordering::Less;
    } else if a > b {
        return Ordering::Greater;
    }
    Ordering::Equal
}

fn main() {

    input!{
        N:usize,
        T:[isize;N],
        A:[isize;N]
    }

    let MOD = 1_000_000_007;
    let mut ans = 1;

    if N==1{
        println!("{}",if A[0]==T[0] {1} else {0});
        return;
    }

    if A[0]>A[1] && A[0]!=T[0] {ans = 0;}
    if A[0]<T[0] {ans = 0;}
    if T[N-1]>T[N-2] && A[N-1]!=T[N-1] {ans = 0;}
    if T[N-1]<A[N-1] {ans = 0;}

    for i in 1..N-1{
        let mut tmp = 1;
        if T[i]>T[i-1] && A[i]>A[i+1]{
            if T[i]!=A[i] {tmp = 0;}
        } else if T[i]>T[i-1]{
            if T[i] > A[i] {tmp = 0;}
        } else if A[i]>A[i+1]{
            if A[i] > T[i] {tmp = 0;}
        } else {
            tmp = min(T[i],A[i]);
        }
        ans *= tmp;
        ans %= MOD;
    }

    ans %= MOD;
    println!("{}",ans);
}