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 <queue>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define int long long
#define Maxn 200000
priority_queue<pair<int,int> >q;
int n,m,b[Maxn+5],a[Maxn+5],pre[Maxn+5],nxt[Maxn+5];
signed main(){
	int i,j,k,ans=0;
	scanf("%lld",&n);
	for(i=1; i<=n; i++)scanf("%lld",&a[i]);
	for(i=1; i<=n; i++) {
		scanf("%lld",&b[i]);
		q.push(mp(b[i],i));
	}
	for(i=1; i<=n; i++) {
		nxt[i]=i+1;
		pre[i]=i-1;
	}
	pre[1]=n;
	nxt[n]=1;
	while(!q.empty()) {
		int x=q.top().se;
		q.pop();
		if(b[x]==a[x])continue;
		int y=pre[x],z=nxt[x];
		if(b[x]-b[y]-b[z]<a[x]) {
			puts("-1");
			return 0;
		}
		k=b[x]-a[x];
		ans+=k/(b[y]+b[z]);
		b[x]=(k%(b[y]+b[z]))+a[x];
		if(a[x]!=b[x])q.push(mp(b[x],x));
	}
	cout<<ans<<"\n";
	return 0;
}
