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 rep(i,n) for(int i=0; i<(n); i++)
const int MX = 200005;
int n, a[MX], b[MX], c[MX], pr, nx;
long long cnt;
stack<int> st;
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%d", a+i);
	rep(i,n){
		scanf("%d", b+i);
		pr = i ? i-1 : n-1;
		nx = i != n-1 ? i+1 : 0;
		c[pr] += b[i];
		c[nx] += b[i];
		b[i] -= a[i];
	}
	rep(i,n) if(b[i] >= c[i]) st.push(i);
	while(!st.empty()){
		int t = st.top(); st.pop();
		int d = b[t] / c[t] * c[t];
		cnt += b[t] / c[t];
		pr = t ? t-1 : n-1;
		nx = t != n-1 ? t+1 : 0;
		c[pr] -= d; if(b[pr] >= c[pr]) st.push(pr);
		c[nx] -= d; if(b[nx] >= c[nx]) st.push(nx);
		b[t] -= d;
	}
	rep(i,n) if(b[i]) return 0*puts("-1");
	printf("%lld\n", cnt);
}