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 <iostream>
#include <vector>
#include <queue>

using namespace std;

long long solve(vector<int> A, vector<int> B){
    const int N = A.size();
    long long res = 0;
    priority_queue<pair<int,int>> qu;
    for(int i=0;i<N;i++){
        if(A[i] > B[i]) return -1;
        if(A[i] == B[i]) continue;
        qu.emplace(B[i], i);
    }
    while(!qu.empty()){
        int idx = qu.top().second; qu.pop();
        int sub = B[(idx+1)%N] + B[(idx+N-1)%N];
        int target = max(A[idx], max(B[(idx+1)%N], B[(idx+N-1)%N]));
        int num = max(1, (B[idx]-target)/sub);
        B[idx] -= num * sub;
        res += num;
        if(B[idx] < A[idx]) return -1;
        if(B[idx] > A[idx]) qu.emplace(B[idx], idx);
    }
    return res;
}

int main(){
    int N;
    while(cin >> N){
        vector<int> A(N), B(N);
        for(auto& t : A) cin >> t;
        for(auto& t : B) cin >> t;
        cout << solve(A, B) << endl;
    }
}