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 <stdio.h>
# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
# define x first
# define y second
# define ll long long
# define IOS ios_base :: sync_with_stdio(0);cin.tie(0)
# define p(v) cerr << #v << " = " << v << '\n'
# define p2(v) cerr << #v << " = " << (complex < int > (v.x,v.y)) << '\n'
# define vi vector < int >
# define vll vector < ll >
# define pii pair < int , int >
# define mp make_pair
const int mod = 1e9 + 7;
int main(void)
{
    #ifdef CF
    freopen("input","r",stdin);
    #endif // CF
    srand(time(0));
    fo << fixed << setprecision(7);
    cerr << fixed << setprecision(7);
    static int a[1 << 20];
    static int b[1 << 20];
    static int A[1 << 20];
    static int B[1 << 20];
    int n;
    fi>>n;
    for (int i = 1;i <= n;++i)
        fi>>a[i];
    for (int i = 1;i <= n;++i)
        fi>>b[i];
    for (int i = 1;i <= n;++i)
        if (a[i] != a[i-1])
            A[i] = a[i];
    for (int i = n;i;--i)
        if (b[i] != b[i+1])
            B[i] = b[i];
    for (int i = 1;i <= n;++i)
        if (A[i] > b[i] || B[i] > a[i] || (A[i] && B[i] && A[i] != B[i]))
            return fo << "0\n",0;
    int ans = 1;
    for (int i = 1;i <= n;++i)
        if (!A[i] && !B[i])
            ans = (1ll * ans * min(a[i],b[i])) % mod;
    fo << ans << '\n';
    cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
    return 0;
}