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 <bits/stdc++.h>
#define REP(i, n) for(int i = 0;i < n;i++)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())
#define ll long long
#define pb(a) push_back(a)
#define INF 1000000000
#define LINF 1e18
#define MOD 1000000007
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

typedef vector<unsigned int>vec;
typedef vector<vec> mat;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const double PI = acos(-1.0);
const double EPS = 1e-10;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin>>N;
    vector<ll> t(N),a(N);
    REP(i,N) cin>>t[i];
    REP(i,N) cin>>a[i];
    if(a[0]!=t[N-1]){
        cout<<0<<endl;
        return 0;
    }
    ll ans = 1;
    for(int i = 1; i < N-1; i++){
		if(t[i] == t[i-1] && a[i] == a[i+1]){
			ans *= min(t[i],a[i]);
			ans %= MOD;
		}else if(t[i] > t[i-1] && a[i] == a[i+1] && a[i] < t[i]){  //h[i] == t[i]
			ans = 0;
		}else if(t[i] == t[i-1] && a[i] > a[i+1] && a[i] > t[i]){  //h[i] == a[i]
			ans = 0;
		}else if(t[i] > t[i-1] && a[i] > a[i+1] && a[i] != t[i]){   //h[i] == t[i] == a[i]
			ans = 0;
		}
	}
    cout<<ans<<endl;
}