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: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cmath>

using namespace std;

#define NNN 1000000007

int main()
{
  int N;
  cin >> N;
  vector<long long int> T(N), A(N);
  long long int MaxT = 0, MaxA = 0;

  for(int i = 0; i < N; i++) {cin >> T[i]; MaxT = max(MaxT, T[i]);}
  for(int i = 0; i < N; i++) {cin >> A[i]; MaxA = max(MaxA, A[i]);}
  if(MaxT != MaxA) {cout << "0" << endl; return 0;}

  vector<long long int> dp(N);

  dp[0] = 1;
  dp[N-1] = 1;

  for(int i = 1; i < N-1; i++) {
    if(T[i-1] != T[i]) dp[i] = 1;
    else dp[i] = T[i]%NNN;
  }
  for(int i = N-2; i >= 0; i--) {
    if(A[i+1] != A[i]) dp[i] = 1;
    else dp[i] = min(A[i], dp[i])%NNN;
  }
  long long int ans = 1;
  //for(auto u: dp) cout << u << endl;
  for(int i = 0; i < dp.size(); i++) {
    ans *= (dp[i]%NNN);
    ans = ans % NNN;
  }
  cout << (ans%NNN) << endl;

  /*
  for(int i = 1; i < N-2; i++) {
    if(T[i-1] != T[i]) dp[i] = dp[i-1] % NNN;
    else if(A[i-1] != A[i]) dp[i] = dp[i-1] % NNN;
    else if(A[i] != A[i+1]) dp[i] = dp[i-1] % NNN;
    else {
      if(T[i] < A[i]) dp[i] = (dp[i-1] * T[i]) % NNN;
      else dp[i] = (dp[i-1] * A[i]) % NNN;
    }
  }
  for(auto u: dp) cout << u<< endl;
  cout << dp[N-3]%NNN << endl;
  */

  return 0;
}