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

import java.util.ArrayDeque;
import java.util.Scanner;


public class Main {
	
	

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int[] begin = new int[n];
		int[] end = new int[n];
		for (int i=0;i<n;i++) {
			begin[i]=sc.nextInt();
		}
		for (int i=0;i<n;i++) {
			end[i]=sc.nextInt();
		}
		
		long count = 0;
		
		ArrayDeque<Integer> check = new ArrayDeque<>();
		
		for (int i=0;i<n;i++) {
			check.offer(i);
		}
		while (!check.isEmpty()) {
			int i = check.poll();
			int times = process(end, begin, i);
			if (times>0) {
				count += times;
				check.offer(i==0 ? n-1 : i-1);
				check.offer(i==n-1 ? 0 : i+1);
			}
		}
		
		boolean same = true;
		for (int i=0;i<n;i++) {
			if (begin[i]!=end[i]) {
				same=false;
				break;
			}
		}
		if (same) {
			System.out.println(count);
		} else {
			System.out.println(-1);
		}
	}
	
	static int process(int[] end, int[] begin, int i) {
		int left = (i==0 ? end[end.length-1] : end[i-1]);
		int right = (i==end.length-1 ? end[0] : end[i+1]);

		if (end[i]>left+right && end[i]>begin[i]) {
			int ret = (end[i]-begin[i])/(left+right);
			end[i]-=((left+right)*ret);
			return ret;
		} else {
			return 0;
		}
	}
}