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<cstdio>
#include<queue>
#define RI register int
#define CI const int&
using namespace std;
const int N=200005;
int n,a[N],b[N],pre[N],nxt[N]; long long ans; queue <int> q; bool vis[N];
inline bool check(CI id)
{
	return b[id]-a[id]>=b[pre[id]]+b[nxt[id]];
}
int main()
{
	RI i; for (scanf("%d",&n),i=1;i<=n;++i) scanf("%d",&a[i]);
	for (i=1;i<=n;++i) scanf("%d",&b[i]),pre[i]=i-1,nxt[i]=i+1;
	for (nxt[pre[1]=n]=i=1;i<=n;++i) if (check(i)) q.push(i),vis[i]=1;
	while (!q.empty())
	{
		int nw=q.front(); vis[nw]=0; q.pop();
		int cur=(b[nw]-a[nw])/(b[pre[nw]]+b[nxt[nw]]);
		ans+=cur; b[nw]-=cur*(b[pre[nw]]+b[nxt[nw]]);
		if (!vis[pre[nw]]&&check(pre[nw])) q.push(pre[nw]),vis[pre[nw]]=1;
		if (!vis[nxt[nw]]&&check(nxt[nw])) q.push(nxt[nw]),vis[nxt[nw]]=1;
	}
	for (i=1;i<=n;++i) if (a[i]!=b[i]) return puts("-1"),0;
	return printf("%lld",ans),0;
}