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;
const int n = 2e5 + 5;

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

int main(){
	cin >> N;
	for(int i = 0; i < N; ++i){
		cin >> A[i];
	}
	priority_queue<pair<int,int>> Q;
	for(int i = 0; i < N; ++i){
		cin >> B[i];
		if(B[i] > A[i])
			Q.push({B[i], i});
	}
	ll c = 0;
	while(!Q.empty()){
		int diff = Q.top().first;
		int i = Q.top().second;
		//cerr << "i: "<< i << " diff: "<< diff << endl;
		Q.pop();
		if(B[i] <= A[i])
			break;
		int nextV = max(B[(i+1)%N], B[(i+N-1)%N]);
		nextV = max(A[i], nextV);
		int nbS = B[(i+1)%N] + B[(i+N-1)%N];
		int cc = max(1, (B[i]-nextV)/nbS);
		B[i] -= cc*nbS;
		c+=cc;
		if(B[i] > A[i])
			Q.push({B[i], i});
		if(B[i] < 0)
			break;
	}
	bool v = true;
	for(int i = 0;i <N; ++i){
		v = v && (A[i] ==B[i]);
	}
	cout << (v?c:-1) << endl;
	return 0;
	
}
