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 FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)
using namespace std;
#define MOD 1000000007

int main()
{
    int N;
    scanf("%d", &N);
    vector<int> T(N), A(N);
    int max = 0, max_index = 0;
    REP (i, N) {
        scanf("%d", &T[i]);
        if (max < T[i]) {
            max = T[i];
            max_index = i;
        }
    }
    REP (i, N) {
        scanf("%d", &A[i]);
    }
    if (A[max_index] != max) {
        printf("0\n");
        return 0;
    }
    if (A[0] != T[N - 1]) {
        printf("0\n");
        return 0;
    }
    if (T.size() == 1) {
        if (T[0] == A[0]) {
            printf("1\n");
            return 0;
        } else {
            printf("0\n");
            return 0;
        }
    }
    long long ans = 1;
    for (int i = 1; i < max_index; i++) {
        if (T[i] == T[i - 1]) {
            ans *= T[i] % MOD;
            ans %= MOD;
        }
    }
    for (int i = N - 2; i > max_index; i--) {
        if (A[i] == A[i + 1]) {
            ans *= A[i] % MOD;
            ans %= MOD;
        }
    }
    printf("%lld\n", ans);
}