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>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
template<class T> using V = vector<T>;
template<class T> using VV = vector<vector<T> >;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define repb(i, n) for(ll i = (n)-1; i >= 0; i--)
#define repr(i, a, b) for(ll i = (a); i < (b); i++)
#define reprb(i, a, b) for(ll i = (a)-1; i >= (b); i--)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
const ll MOD = 1000000007;
const ll INF = 100000000000000000LL;
inline ll GCD(ll a, ll b){ return b?GCD(b, a % b):a; }
inline ll LCM(ll a, ll b){ return a/GCD(a, b)*b; }
inline ll powint(unsigned long long x, ll y){ ll r=1; while(y){ if(y&1) r*=x; x*=x; y>>=1; } return r; }
inline ll powmod(ll x, ll y, ll m = MOD){ ll r=1; while(y){ if(y&1) r*=x; x*=x; r%=m; x%=m; y>>=1; } return r; }
template<class T>bool chmax(T &a, const T &b){ if(a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b){ if(b<a) { a=b; return 1; } return 0; }
#ifdef OJ_LOCAL
#include "dump.hpp"
#else
#define dump(...) ((void)0)
#endif

void pr(ll n){
    cout << n << endl;
    exit(0);
}

int main(){
    cin.tie(0); ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll n;
    cin >> n;
    vll t(n);
    rep(i, n){
        cin >> t[i];
    }
    vll a(n);
    rep(i, n){
        cin >> a[i];
    }
    vll ans1(n, INF);
    vll ans2(n, INF);
    vi ansd1(n, 0);
    vi ansd2(n, 0);
    rep(i, n){
        if(i == 0 || t[i] > t[i-1]){
            chmin(ans1[i], 1ll);
            ansd1[i] = 1;
        }else{
            chmin(ans1[i], t[i]);
        }
    }
    repb(i, n){
        if(i == n-1 || a[i] > a[i+1]){
            chmin(ans2[i], 1ll);
            ansd2[i] = 1;
        }else{
            chmin(ans2[i], a[i]);
        }
    }
    rep(i, n){
        if(ansd1[i] == 1 && ansd2[i] == 1 && t[i] != a[i]){
            pr(0);
        }
        if(ansd1[i] == 1 && ansd2[i] == 0 && t[i] > a[i]){
            pr(0);
        }
        if(ansd1[i] == 0 && ansd2[i] == 1 && t[i] < a[i]){
            pr(0);
        }
    }
    ll ansm = 1;
    rep(i, n){
        ansm *= min(ans1[i], ans2[i])%MOD;
        ansm %= MOD;
    }
    cout << ansm << endl;
    return 0;
}