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 mpr make_pair
typedef pair < int , int > pii;
typedef long long LL;
int n,a[200005],b[200005];
multiset < pii > st;
LL res;
void dec(int d){
	st.erase(st.find(mpr(b[d],d)));
	int S=b[(d+n-1)%n]+b[(d+1)%n];
	if(S>b[d]){
		printf("-1\n");
		exit(0);
	}
	int e=(b[d]-a[d])/S;
	if(!e){
		printf("-1\n");
		exit(0);
	}
	res+=(LL)e;
	b[d]-=(LL)S*(LL)e;
	if(b[d]<a[d]){
		printf("-1\n");
		exit(0);
	}
	if(b[d]>a[d]) st.insert(mpr(b[d],d));
}
int main(){
	int i,j,k;
	scanf("%d",&n);
	for(i=0;i<n;++i) scanf("%d",a+i);
	for(i=0;i<n;++i) scanf("%d",b+i);
	for(i=0;i<n;++i){
		if(b[i]>a[i]){
			st.insert(mpr(b[i],i));
		}
		else if(b[i]<a[i]){
			printf("-1\n");
			return 0;
		}
	}
	while(!st.empty()){
		dec(st.rbegin()->second);
	}
	for(i=0;i<n;++i){
		if(a[i]!=b[i]){
			printf("-1\n");
			return 0;
		}
	}
	printf("%lld\n",res);
	return 0;
}