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 N 200200

int n, a[N], b[N];

long long solve() {
    long long rlt = 0;
    while (1) {
        bool fg = 1;
        for (int i = 1; i <= n; i ++) {
            if (a[i] > b[i]) return -1;
            if (a[i] == b[i]) continue;
            int k = b[i>1?i-1:n] + b[i<n?i+1:1];
            if (b[i] <= k) continue;
            if (b[i] - a[i] < k) return -1;
            fg = 0;
            rlt += (b[i] - a[i]) / k;
            b[i] = (b[i] - a[i]) % k + a[i];
        }
        if (fg) break;
    }
    if (!rlt) return -1;
    return rlt;
}

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]);
    printf("%lld\n", solve());
}
