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 <cassert>
#include <cctype>
#include <cmath>
#include <climits>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <bitset>
#include <deque>
#include <functional>
#include <iterator>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <fstream>
#include <iostream>
#include <sstream>
#include <iomanip>

using namespace std;
typedef pair<int, int> pint;
typedef vector<int> vint;
typedef long long ll;
#define ALL(c) (c).begin(),(c).end()
#define RALL(c) (c).rbegin(),(c).rend()
#define SORT(c) sort(ALL(c))
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define NEXTP(c) next_permutation(ALL(c))
#define SUM(c) accumulate(ALL(c),0)
#define MP( a, b ) make_pair( ( a ), ( b ) )
#define ANS(x) cout << (x) << endl;
#define DUMP(x) cerr << #x << " = " << (x) << endl;
#define INF 1000000000 // 10^9
const double EPS = 1e-10;

int main(){
    int N;
    cin >> N;

    int T[N], A[N];
    REP(i,N)cin>>T[i];
    REP(i,N)cin>>A[i];

    int p[N] = {};
    p[0] = p[N-1] = 1;

    FOR(i,1,N-1){
        if(T[i]>T[i-1] || A[i]>A[i+1]){
            if (T[i]>T[i-1] && T[i] > A[i]) p[i] = 0;
            else if (A[i]>A[i+1] && A[i] > T[i]) p[i] = 0;
            else p[i] = 1;
        }else{
            p[i] = min(T[i], A[i]);
        }
    }

    ll ret = 1;
    REP(i,N) ret = (ret * p[i]) % 1000000007;

    cout << ret << endl;

    return 0;
}