#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++(i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++(i))
#define REP_R(i, n) for (int i = (int)(n)-1; (i) >= 0; --(i))
#define REP3R(i, m, n) for (int i = (int)(n)-1; (i) >= (int)(m); --(i))
#define ALL(x) ::std::begin(x), ::std::end(x)
using namespace std;

template <typename T>
void print_vector(T& vec) {
    stringstream ss;
    //ss << "{";
    for (auto v : vec) ss << v << " ";
    //ss << "}" << endl;
    cout << ss.str() << endl;
}

template <typename T>
void print_pairs(T& vec) {
    stringstream ss;
    ss << "{";
    for (auto v : vec) ss << v.first << "," << v.second << " ";
    ss << "}" << endl;
    cout << ss.str() << endl;
}

template <typename T>
string string_vector(vector<T>& vec) {
    stringstream ss;
    //ss << "{";
    for (auto v : vec) ss << v << endl;
    //ss << "}\n";
    return ss.str();
}

template <typename T>
string string_vector(vector<vector<T>>& vec) {
    stringstream ss;
    //ss << "{";
    for (auto v : vec) {
        //ss << "{";
        for (auto vv : v) ss << vv << " ";
        //ss << "}";
    }
    //ss << "}\n";
    return ss.str();
}

using ll = long long;
class Solution {
   public:
    vector<int> seen;
    vector<int> loc;
    
    vector<int> solve(int N, vector<ll>& P) {        
        int val = 1, pos = 0, start;
        vector<int> res;
        
        for(int i = 1; i <=N ; ++i) {
            if(loc[i] == i ) continue;
            for (int j = loc[i] - 1; j >= i; j--) {
                if(seen[j]++) {
                    res.clear();
                    return res;
                }
                
                loc[P[j]] = j+1;
                loc[P[j+1]] = j;
                swap(P[j], P[j+1]);
                res.push_back(j);
            }
        }
        return res;
    }
};

// generated by online-judge-template-generator v4.7.1 (https://github.com/online-judge-tools/template-generator)
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    Solution sol;
    int N;
    cin >> N;
    vector<ll> P(N+1);
    sol.loc.resize(N+1);
    sol.seen.resize(N+1);
    REP (i, N) { cin >> P[i+1]; sol.loc[ P[i+1] ] = i+1;}
    vector<int> ans = sol.solve(N, P);
    if(ans.size() != N - 1) cout << -1 << endl;
    else cout << string_vector(ans) << endl;
    return 0;
}


#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>   
using namespace std;
using namespace __gnu_pbds;
typedef long long ll; 
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll> > vv;
typedef vector<string>  vs;
typedef vector<pair<ll,ll>> vpr;
typedef vector<bool>vb;
typedef vector<double>vd;
typedef long double ld;
#define f(i,n) for(ll i = 0; i < n; i++)
#define ff(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define amax(x,y) if(y>x) x=y;
#define amin(x,y) if(y<x)x=y;
#define bg(x) x.begin()
#define sz(x) (ll)x.size()
#define in(x,n) for(ll i=0;i<n;i++)cin>>x[i]
#define out(x,n) for(ll i=0;i<n;i++)cout<<x[i]<<" " 
#define mxt(a) *(max_element(a.begin(),a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc ll t;cin>>t;while(t--)
typedef pair<ll,ll> pi;
#define yes cout<<"YES\n";
// #define no cout<<"NO\n";
// #define yesno(f) if(f) yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1,ll y1){if((x1%y1)==0)return x1/y1;else return x1/y1+1;}
ll power(ll a,ll b,ll m)
{
    if(b==0)
        return 1;
    ll d=power(a,b/2,m);
    d=(d*d)%m;
    if(b&1)
        d=(d*a)%m;
    return d;
}
const ll mod=1e9+7;
int MOD(int a)
{
    if(a<0)
        a+=mod;
    if(a>=mod)
        a%=mod;
    return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time
// set_name.order_of_key(k) It returns to the number of items that are strictly smaller than our item k in O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
void no()
{
    cout<<-1<<endl;
    exit(0);
}
int32_t main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    int n;cin>>n;
    vector<int>a(n);in(a,n);
    vector<int>b(n);in(b,n);
    vector<int>p(n);in(p,n);
    vector<int>inv(n);
    for (int i = 0; i < n; ++i)
    {
        p[i]--;
        inv[p[i]]=i;
    }

    // return 0;   
    vpr operations;
    vb completed(n,false);
    for(int i=0;i<n;i++)
    {
        if(completed[i])
            continue;
        int u=i;
        vector<int>cycle;
        while(p[u]!=i)
        {
            cycle.pb(u);
            completed[u]=true;
            u=p[u];
        }
        // continue;
        cycle.pb(u);
        completed[u]=true;
        if(sz(cycle)!=1)
        {
            priority_queue<pi,vector<pi>,greater<pi> > que;
            for(auto it:cycle)
            {
                que.push({a[it],it});
            }
            for(int j=0;j<cycle.size()-1;j++)
            {
                int lightest_i=que.top().second;
                que.pop();
                if(b[p[lightest_i]] >= a[lightest_i])
                    no();
                int a=inv[lightest_i];
                operations.pb({lightest_i,a});
                int b=p[lightest_i];
                inv[b]=a;
                p[a]=b;
            }
        }
    }
    cout<<operations.size()<<endl;
    for(auto it:operations)   
        cout<<it.first+1<<" "<<it.second+1<<endl;
    return 0;
}