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 sz(x) ((int) (x).size())

typedef long long ll;

const int n = 2e5+10;

int N;
ll A[n], B[n];

int main(){
	cin >> N;
	for(int i=0; i<N; i++){
		cin >> A[i];
	}
	for(int i=0; i<N; i++){
		cin >> B[i];
	}
	ll cnt = 0;
	while(true){
		bool f = true;
		bool changed = false;
		for(int i=0; i<N; i++){
			ll sum = B[(i+N-1)%N]+B[(i+1)%N];
			if(B[i] >= sum){
				if((B[i]-A[i]) % sum == 0){
					cnt += (B[i]-A[i])/sum;
					if(B[i] != A[i])
						changed = true;
					B[i] = A[i];
				}
				else{
					cnt += B[i]/sum;
					changed = true;
					B[i] %= sum;
				}
				if(B[i] < A[i]){
					cout << -1 << endl;
					return 0;
				}
			}
			if(B[i] != A[i])
				f = false;
		}
		if(f){
			cout << cnt << endl;
			break;
		}
		if(!changed){
			cout << -1 << endl;
			break;
		}
	}

  return 0;
}
