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>
#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <queue>

#define rep(i,n) for(int i=0;i<n;i++)
#define dup(x,y) (((x)+(y)-1)/(y)) // dup * y >= x なる最小のdup.
using namespace std;
typedef long long ll;

int main() {
	ll N;
	cin >> N;
	vector<ll> A(N), B(N);
	rep(i, N) cin >> A[i];
	rep(i, N) cin >> B[i];

	priority_queue<pair<ll, ll>> task; // <B[i], i>;
	rep(i, N) {
		if(A[i] < B[i]) {
			task.push(make_pair(B[i], i));
		}
	}
	
	ll ans = 0;
	while(task.size()>0) {
		ll id = task.top().second;
		task.pop();
		ll s = B[(id+N-1)%N] + B[(id+1)%N];
		if(B[id]-A[id]>=s){
			ll op = (B[id]-A[id])/s;
			ans += op;
			B[id] -= op*s;
			if(A[id]<B[id]) {
				task.push(make_pair(B[id], id));
			}
		} else {
			cout << -1 << endl;
			return 0;
		}
	}
	
	cout << ans << endl;
	return 0;
}