#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep_r(i,a,b) for(int i=a;i>=b;i--)
#define each(a,x) for (auto& x : a)
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define sz(x) int(x.size())
#define so(x) sort(all(x))
#define so_r(x) sort(all(x),greater<int>())
#define lb lower_bound
#define ub upper_bound
const char nl = '\n';
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int bit_cnt(int x){
    return __builtin_popcount(x);
}
ll bex(ll a, ll b, ll mod = 1e9 + 7){ll res = 1LL; while(b){ if (b&1) res = res * a % mod; a = a * a % mod; b >>= 1;} return res;}
template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}

const int MOD = 998244353;
const int N = 5e5 + 5;
const ll INF = 1e16;
// https://yukicoder.me/problems/no/1120
// Brief: N students, real points are A1, A2, ... AN 
// but teacher wants to change to B1, B2, ... BN
// in each operation, teacher can increase 1 point for one student but decrease all other students 1 point 
// find out minimum operations to achieve that, if impossible print -1

int n; 
int a[N], b[N];
void solve(){
    cin >> n;
    rep(i,1,n) cin >> a[i];
    rep(i,1,n) cin >> b[i];
    ll dif = 0;
    rep(i,1,n) dif+=a[i]-b[i];

    // each operation will make total points decrease n-2
    // => dif % (n-2) == 0
    // and dif > 0
    
    // special case is n == 2
    // => dif == 0
    if (n == 2){
        if (dif != 0) cout << -1 << nl;
        cout << abs(a[1] - b[1]) << nl; 
        return;
    }

    if (dif % (n-2) != 0 || dif < 0){
        cout << -1 << nl;
        return;
    }

    ll operations = dif / (n-2); // number of operations will be used
    // add[i] is the number of operations teacher use for increasing point for student[i]
    // b[i] = a[i] - operations + 2 * add[i] 
    vl add(n+1);
    rep(i,1,n){
        ll tmp = b[i] - a[i] + operations; 
        if (tmp % 2 == 1 || tmp < 0) {
            cout << -1 << nl;
            return;
        }
        add[i] = tmp / 2;
    }

    ll tot = operations;
    rep(i,1,n){ 
        tot -= add[i];
    }

    if (tot != 0) {
        cout << -1 << nl;
    } else {
        cout << operations << nl;
    }
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t = 1; 
    // cin >> t;

    while (t--){
        solve();
    }
}