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.

Examples

Input

5
1 3 3 3 3
3 3 2 2 2


Output

4


Input

5
1 1 1 2 2
3 2 1 1 1


Output

0


Input

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5


Output

884111967


Input

1
17
17


Output

1
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;
using ll = long long;
static const int MOD = 1000000007;

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n), b(n);
	for(int i = 0; i < n; ++i){ cin >> a[i]; }
	for(int i = 0; i < n; ++i){ cin >> b[i]; }
	vector<bool> fixed(n);
	fixed[0] = fixed[n - 1] = true;
	int a_max = 0, b_max = n - 1;
	for(int i = 1; i < n; ++i){
		if(a[i] > a[i - 1]){
			fixed[i] = true;
			a_max = i;
		}
	}
	for(int i = n - 2; i >= 0; --i){
		if(b[i] > b[i + 1]){
			fixed[i] = true;
			b_max = i;
		}
	}
	if(a_max >  b_max || a[a_max] != b[b_max]){
		cout << 0 << endl;
		return 0;
	}
	ll answer = 1;
	for(int i = 0; i < n; ++i){
		if(fixed[i]){ continue; }
		answer = (answer * min(a[i], b[i])) % MOD;
	}
	cout << answer << endl;
	return 0;
}

