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> 
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define ll long long
#define fi first
#define se second
#define PB push_back
#define uint unsigned
#define ull unsigned ll 
using namespace std;
const int N=200005;
int n,a[N],b[N];
ll ans;
priority_queue<pii> Q;
int main(){
	scanf("%d",&n);
	For(i,1,n) scanf("%d",&a[i]);
	For(i,1,n) scanf("%d",&b[i]);
	For(i,1,n) Q.push(pii(b[i],i));
	while (!Q.empty()){
		int x=Q.top().se; Q.pop();
		if (a[x]>b[x]) return puts("-1"),0;
		if (a[x]==b[x]) continue;
		int pre=(x==1?n:x-1);
		int suf=(x==n?1:x+1);
		int S=b[pre]+b[suf];
		int times=(b[x]-a[x])/S;
		if (times==0) return puts("-1"),0;
		ans+=times; b[x]-=times*S;
		if (b[x]<a[x]) return puts("-1"),0;
		Q.push(pii(b[x],x)); 
	} 
	printf("%lld\n",ans);
}