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>

using namespace std;
using ll = long long;

#define rep(i,n) for(int i=0;i<(n);++i)
#define in(v) cin>>v;
#define out(v) cout<<v<<"\n"
template<typename Iterator>void print(const Iterator &s, const Iterator &t) { for(auto it=s;it+1!=t;++it)cout<<*it<<" ";cout<<*(t-1)<<"\n"; }
const int MOD = 1e9+7;

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int N; in(N);

  int T[N], A[N];
  int h[N];
  fill(h, h+N, -1);
  rep(i, N) {
    in(T[i]);
    if (i == 0 || T[i] > T[i-1])
      h[i] = T[i];
  }
  rep(i, N) in(A[i]);

  if (h[N-1] != -1 && h[N-1] != A[N-1]) {
    out(0);
    return 0;
  }
  h[N-1] = A[N-1];
  rep(i, N-1) {
    if (A[i] > A[i+1]) {
      if (h[i] == -1 || h[i] == A[i])
        h[i] = A[i];
      else {
        out(0);
        return 0;
      }
    }
  }
  rep(i, N) {
    if (h[i] != -1) {
      if (h[i] > T[i] || h[i] > A[i]) {
        out(0);
        return 0;
      }
    }
  }
  ll ans = 1;
  rep(i, N) {
    if (h[i] == -1)
      ans = (ll)(ans * min(T[i], A[i])) % MOD;
  }
  out(ans);

  return 0;
}