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 <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define countof(array) (sizeof(array) / sizeof(array[0]))
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,n,s) for(int i = s; i < (n); ++i)
#define rsrep(i,n,s) for(int i = (n)-1; i >= s; --i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define aall(a) (a), (a)+countof(a)//for array sorting
#define raall(a) (a), (a)+countof(a), greater<>()
#define show(x) cout<<#x<<" = "<<x<<endl;
#define vfind(v, a) find(all(v), a) != v.end()
#define yn(f) { if (f) puts("YES"); else puts("NO"); }
#define yns(f) { if (f) puts("Yes"); else puts("No"); }
#define show_ary(...) { cout<<#__VA_ARGS__<<" = "; for (const auto& x : (__VA_ARGS__)) { cout<<x<<" "; } cout<<endl; }
#define show_pair(...) cout<<#__VA_ARGS__<<" = "<<endl; for (const auto& x : (__VA_ARGS__)) { cout<<"  "<<x.fi<<" : "<<x.se<<endl; }
#define out_ary(...) { for (const auto& x : (__VA_ARGS__)) { cout<<x<<" "; } cout<<endl; }
#define argmax(v) distance((v).begin(), max_element(all(v)))
#define argmin(v) distance((v).begin(), min_element(all(v)))
#define vmax(v) *max_element(all(v))
#define vmin(v) *min_element(all(v))
typedef long long int ll;
typedef pair<int, int> P;
typedef vector<P> vpair;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<double> vdouble;
typedef vector<string> vstr;
typedef vector<bool> vbool;
typedef vector<vint> vvint;
typedef vector<vll> vvll;
typedef vector<vstr> vvstr;
typedef vector<vbool> vvbool;
const ll LINF = 2000000000000000000ll;
const int INF = 1000000100;
const ll MOD = 1e9+7;

int main() {
    int n;
    cin >> n;
    vint t(n), a(n);
    rep(i, n) cin >> t[i];
    rep(i, n) cin >> a[i];
    vint mn(n, 0);
    vint ls(n);
    ls[0] = 1;
    mn[0] = t[0];
    rep(i, n-1) {
        if (t[i+1] > t[i] || t[i+1] == 1) {
            ls[i+1] = 1;
            mn[i+1] = t[i+1];
        }
        else ls[i+1] = t[i];
    } 
    ls[n-1] = 1;
    mn[n-1] = a[n-1];
    rrep(i, n-1) {
        if (a[i] > a[i+1] || a[i] == 1) {
            ls[i] = 1;
            if (mn[i] != 0 && mn[i] < a[i]) {
                puts("0");
                return 0;
            }
            mn[i] = a[i];
        }
        else ls[i] = min(ls[i], a[i]);
    } 
    ll ans = 1;
    rep(i, n) {
        ans *= ls[i];
        ans %= MOD;
    }
    if (t[n-1] != a[0]) ans = 0;

    cout << ans << endl;
    return 0;
}