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 int long long

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

const int N = 2e5 + 5;
int a[N], b[N];
priority_queue<pair<int, int> > Q;

signed main()
{
	int n = read();
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i <= n; ++i) b[i] = read(), Q.push({b[i], i});
	
	int ans = 0;
	while (!Q.empty())
	{
		int i = Q.top().second;
		Q.pop();
		if (b[i] == a[i]) continue;
		
		int pre = i == 1 ? n : i - 1;
		int nxt = i == n ? 1 : i + 1;
		if (b[i] - b[pre] - b[nxt] < a[i]) return puts("-1"), 0;
		
		ans += (b[i] - a[i]) / (b[pre] + b[nxt]);
		b[i] = a[i] + (b[i] - a[i]) % (b[pre] + b[nxt]);
		if (a[i] != b[i]) Q.push({b[i], i});
	}
	printf("%lld\n", ans);
	return 0;
}
