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 nn 300008
#define int long long 
#define mod 998244353
int ai[nn],bi[nn];int n;
signed main() {
	#define a ai
	#define b bi
	scanf("%lld",&n);for(int i=1;i<=n;i++) scanf("%lld",&ai[i]);for(int i=1;i<=n;i++) scanf("%lld",&bi[i]);
	#define pii pair<int,int>
	#define mp make_pair
	set<pii> s;for(int i=1;i<=n;i++) s.insert(mp(bi[i],i));
	int ans=0;
	while(s.size()){
		int x=(*s.rbegin()).second;s.erase(--s.end());
		if(b[x]==a[x]) continue;
		if(b[x]<a[x]) break;
		b[0]=b[n],b[n+1]=b[1];
		int su=b[x-1]+b[x+1],dif=b[x]-a[x];
		if(dif<su) break;
		ans+=dif/su,dif%=su;
		b[x]=a[x]+dif;
		s.insert(mp(b[x],x));
	}
	
	for(int i=1;i<=n;i++) if(ai[i]!=bi[i]) return puts("-1")*0;
	cout<<ans;
	return 0;
}

