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 mp(a, b) make_pair(a, b)
typedef pair<int, int> pii;
const int maxn = 2e5 + 10;
int n;
int a[maxn], b[maxn];
typedef long long ll;
ll ans;
priority_queue<pii> Q;
inline int pre(int x) { return x == 1 ? n : x - 1;}
inline int nxt(int x) { return x == n ? 1 : x + 1;}
void no() { puts("-1"); exit(0);}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	for(int i = 1; i <= n; ++i) 
	{
		scanf("%d", &b[i]); 
		if(b[i] > a[i]) Q.emplace(b[i], i);
		if(b[i] < a[i]) no();
	}
	while(!Q.empty())
	{
		pii x = Q.top(); Q.pop();
		int i = x.second;
		int sum = b[pre(i)] + b[nxt(i)];
		if(b[i] < sum) no();
		if((b[i] - a[i]) % sum == 0)
		{
			ans += (b[i] - a[i]) / sum;
			b[i] = a[i];
			continue;
		}
		int step = b[i] / sum;
		b[i] -= step * sum;
		ans += step;
		if(b[i] < a[i]) no();
		Q.emplace(b[i], i);
	}
	printf("%lld\n", ans);
	return 0;
}
			
	
	