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;
 
typedef long long ll;
typedef pair<int, int> ii;

const int N = 2e5+100;

int a[N], b[N];

signed main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
	int n;cin >> n;
	set<ii> st;
	for(int i=0 ; i<n ; i++)
		cin >> a[i];
	for(int i=0 ; i<n ; i++){
		cin >> b[i];
		st.insert(ii(b[i], i));
	}
	ll ans=0;
	while(!st.empty()){
		ii v = *st.rbegin();
		st.erase(v);
		int cur = v.second;
		int nxt = (v.second+1)%n;
		int prv = (v.second-1+n)%n;
		if(b[cur] == a[cur])
			continue;
		int s = b[prv] + b[nxt];
		if(b[cur] < a[cur] || b[cur] < s){
			cout << "-1\n";
			return 0;
		}
		if(a[cur] < s){
			ans += b[cur]/s;
			b[cur] %= s;
			st.insert(ii(b[cur], cur));
			continue;
		}
		if(b[cur] - a[cur] < 0 || (b[cur] - a[cur]) % s != 0){
			cout << "-1\n";
			return 0;
		}
		ans += (b[cur] - a[cur]) / s;
		b[cur] = a[cur];
	}
	cout << ans << "\n";
 	return 0;
}














