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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <functional>
using namespace std;

#define Rep(b, e, i) for(int i = b; i <= e; i++)
#define Repr(e, b, i) for(int i = e; i >= b; i--)
#define rep(n, i) Rep(0, n-1, i)
#define repr(n, i) Repr(n-1, 0, i)
#define all(v) (v).begin(), (v).end()
#define pb(v) push_back(v)
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define bitcnt(x) __builtin_popcount(x)

typedef long long ll;
typedef pair<ll, ll> llP;
typedef pair<int, int> intP;
typedef std::priority_queue<int> IntPrioQueue; //Z->A
typedef std::priority_queue<int, std::vector<int>, std::greater<int> > IntReversePrioQueue; //A->Z

//vector の中身を出力
template <class T>ostream &operator<<(ostream &o,const vector<T>&v)
{o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}

const int MAX = 200000;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<29;
const ll MOD = 1000000007LL;

void solve(void){
    int N;
    cin >> N;
    vector <ll> ts(N), as(N), hs(N, INF);
    rep(N, i) cin >> ts[i];
    rep(N, i) cin >> as[i];
    if (ts[N-1] != as[0]) {
        cout << 0 << endl;
        return;
    }
    hs[0] = ts[0], hs[N-1] = as[N-1];
    reverse(all(as));

    vector <bool> pre(N);
    pre[0] = pre[N-1] = true;

    rep(N-1, i) {
        if (ts[i] != ts[i+1]) {
            hs[i+1] = ts[i+1];
            pre[i+1] = true;
        }
        else hs[i+1] = min(ts[i+1], hs[i+1]);
    }
    reverse(all(hs));
    reverse(all(pre));
    rep(N-1, i) {
        if (as[i] != as[i+1]) {
            if (pre[i+1] && as[i+1] != hs[i+1]) {
                cout << 0 << endl;
                return;
            }
            hs[i+1] = as[i+1];
            pre[i+1] = true;
        }
        else hs[i+1] = min(as[i+1], hs[i+1]);
    }
    /*
    reverse(all(hs));
    cout << hs << endl;
    */
    ll ans = 1LL;
    rep(N, i) {
        if (!pre[i]) {
            ans *= hs[i];
            ans %= MOD;
        }
    }

    cout << ans << endl;

}

int main(void){
  solve();
  //cout << "yui(*-v・)yui" << endl;
  return 0;
}