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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <chrono>
#include <random>
#include <functional>

using namespace std;

typedef long long ll;

const int MOD = 1e9 + 7;
const int INF = 1e9 + 239;

int mul(int a, int b) {
	return ((ll)a * b) % MOD;
}

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	int n;
	cin >> n;
	vector<int> f(n);
	vector<int> s(n);
	for (auto &t : f) {
		cin >> t;
	}
	for (auto &t : s) {
		cin >> t;
	}
	vector<int> l(n, 1);
	vector<int> r(n, INF);
	{
		int c = 0;
		for (int i = 0; i < n; i++) {
			if (c < f[i]) {
				l[i] = f[i];
				r[i] = f[i];
				c = f[i];
			} else {
				r[i] = c;
			}
		}
	}
	{
		int c = 0;
		for (int i = n - 1; i >= 0; i--) {
			if (c < s[i]) {
				l[i] = max(l[i], s[i]);
				r[i] = min(r[i], s[i]);
				c = s[i];
			} else {
				r[i] = min(r[i], c);
			}
		}
	}
	int ans = 1;
	for (int i = 0; i < n; i++) {
		if (l[i] >= r[i]) {
			cout << 0 << endl;
			return 0;
		}
		ans = mul(ans, r[i] - l[i] + 1);
	}
	cout << ans << endl;
}