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 <iostream>
#include <string>
#include <vector>
#include <climits>
#include <set>
#include <algorithm>
#include <unordered_map>
#include <queue>
#include <iomanip>
#include <map>
#include <utility>
#include <numeric>

#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define P pair<int, int>
#define debug(x) cout << #x << ": " << x << ", "
#define debugln(x) cout << #x << ": " << x << '\n'

using namespace std;
using ll = long long;
const int mod = 1e9+7;

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

  int n; cin >> n;

  vector<int> mi(n);
  vector<int> ma(n);

  rep(i, n) {
    int t; cin >> t;
    if (i==0 || t > ma.at(i-1)) {
      mi.at(i) = t;
    } else {
      mi.at(i) = 1;
    }
    ma.at(i) = t;
    //debug(i), debugln(ma.at(i));
  }

  vector<int> a(n);
  rep(i, n) {
    cin >> a.at(i);
  }

  if (a.at(0) != ma.at(n-1)) {
    cout << 0 << endl;
    return 0;
  }

  for(int i=n-1; i>=0; i--) {
    if (i==n-1 || a.at(i) > a.at(i+1)) {
      //debug(i); debug(a.at(i)); debugln(ma.at(i));
      if (ma.at(i) < a.at(i)) {
        cout << 0 << endl;
        return 0;
      }
      mi.at(i) = a.at(i);
    } else {
      mi.at(i) = max(mi.at(i), 1);
    }
    ma.at(i) = min(ma.at(i), a.at(i));
  }

  ll ans = 1;
  rep(i, n) {
    ans = ans * (ma.at(i) - mi.at(i) + 1) % mod;
  }
  cout << ans << endl;
}