There are N positive integers arranged in a circle.

Now, the i-th number is A_i. Takahashi wants the i-th number to be B_i. For this objective, he will repeatedly perform the following operation:

* Choose an integer i such that 1 \leq i \leq N.
* Let a, b, c be the (i-1)-th, i-th, and (i+1)-th numbers, respectively. Replace the i-th number with a+b+c.



Here the 0-th number is the N-th number, and the (N+1)-th number is the 1-st number.

Determine if Takahashi can achieve his objective. If the answer is yes, find the minimum number of operations required.

Constraints

* 3 \leq N \leq 2 \times 10^5
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N
B_1 B_2 ... B_N


Output

Print the minimum number of operations required, or `-1` if the objective cannot be achieved.

Examples

Input

3
1 1 1
13 5 7


Output

4


Input

4
1 2 3 4
2 3 4 5


Output

-1


Input

5
5 6 5 2 1
9817 1108 6890 4343 8704


Output

25
#include <bits/stdc++.h>
using namespace std;
						
#define fast_cin() ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl '\n'
#define int long long
				
int32_t main() {
	fast_cin();  
	int n; cin >> n;
	vector<int> a(n), b(n);
	set<pair<int, int> > st;
	for (int i = 0; i < n; ++i) cin >> a[i];
	for (int i = 0; i < n; ++i) {
		cin >> b[i];
		st.insert({b[i], i});
	}
	int ans = 0;
	while (!st.empty()) {
		int i = st.rbegin()->second;
		st.erase(--st.end());
		int step = b[(i + 1) % n] + b[(i - 1 + n) % n];
		if (b[i] < step and b[i] != a[i]) {
			return cout << -1, 0;
		}
		if ((b[i] - a[i]) % step == 0) {
			ans += (b[i] - a[i]) / step;
			b[i] = a[i];
			continue;
		}
		ans += (b[i] / step);
		b[i] %= step;
		if (b[i] < a[i]) {
			return cout << -1, 0;
		}
		st.insert({b[i], i});
	}
	cout << ans << endl;
	return 0;
}