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>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
const int MAXN = 200005;
using lint = long long;

int n, a[MAXN], b[MAXN];
queue<int> que;

int opable(int x){
	int l = (x + n - 1) % n;
	int r = (x + 1) % n;
	return max(0, b[x] - a[x]) / (b[l] + b[r]);
}

void solve(){
	scanf("%d",&n);
	for(int i=0; i<n; i++) scanf("%d",&a[i]);
	for(int i=0; i<n; i++) scanf("%d",&b[i]);
	for(int i=0; i<n; i++){
		if(opable(i)) que.push(i);
	}
	lint ans = 0;
	while(sz(que)){
		auto x = que.front(); que.pop();
		int l = (x + n - 1) % n, r = (x + 1) % n;
		ans += opable(x);
		b[x] -= (b[l] + b[r]) * opable(x);
		if(opable(l)) que.push(l);
		if(opable(r)) que.push(r);
	}
	for(int i=0; i<n; i++){
		if(a[i] != b[i]){
			puts("-1");
			return;
		}
	}
	cout << ans << endl;
}

int main(){
	solve();
}
