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;

const int _ = 2e5 + 3;
int A[_] , B[_] , N;
struct cmp{bool operator ()(int x , int y){return A[x] > A[y];}};
multiset < int , cmp > pos;

int main(){
	ios::sync_with_stdio(0); cin >> N;
	for(int i = 1 ; i <= N ; ++i) cin >> B[i];
	for(int i = 1 ; i <= N ; ++i){cin >> A[i]; pos.insert(i);}
	long long stp = 0;
	while(!pos.empty()){
		int t = *pos.begin(); pos.erase(pos.begin());
		if(A[t] < B[t]){puts("-1"); return 0;}
		if(A[t] > B[t]){
			int num = A[t == 1 ? N : t - 1] + A[t == N ? 1 : t + 1];
			if(A[t] % num == B[t] % num){stp += (A[t] - B[t]) / num; A[t] = B[t];}
			else{
				if(A[t] < num){puts("-1"); return 0;}
				stp += A[t] / num; A[t] %= num;
				if(A[t] < B[t]){puts("-1"); return 0;}
				pos.insert(t);
			}
		}
	}
	printf("%lld\n" , stp);
	return 0;
}
