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 fi first 
#define se second
using namespace std;
typedef long long i64;
typedef pair<int, int> pii;
const int N = 2e5+5;
int n, a[N], b[N]; i64 ans; priority_queue<pii> q;
int get(int x) { return !x?n:x==n+1?1:x; }
int main()
{ 
    scanf("%d", &n); for(int i = 1; i <= n; ++i) scanf("%d", a+i);
    for(int i = 1; i <= n; ++i) scanf("%d", b+i);
    for(int i = 1; i <= n; ++i) 
        if(a[i] > b[i]) return puts("-1"), 0;
        else if(a[i] < b[i]) q.push(pii(b[i], i));
    while(q.size())
    {
        pii u = q.top(); q.pop();
        int i = u.se, d = b[i]-a[i], w = b[get(i-1)]+b[get(i+1)];
        if(d/w == 0&&a[i] != b[i]) return puts("-1"), 0;
        ans += d/w, b[i] -= d/w*w;
        if(a[i] < b[i]) q.push(pii(b[i], i));
    }
    printf("%lld\n", ans);
    return 0;
}