#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
//data structures
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<ll> vll;
typedef vector<ld> vld;
typedef pair<long long, long long> pll;
typedef pair<char, int> ci;
typedef pair<string, int> si;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define sz(a) ((int)a.size())
#define fi first
#define se second
#define whole(v) v.begin(), v.end()
#define rwhole(v) v.rbegin(), v.rend()
#define fro front
#define pqueue priority_queue
#define ubound upper_bound
#define lbound lower_bound
#define beg(v) v.begin()
//bit operations
int flip(int x){
return ~(x) ^ (1 << 32);
}
int allon(int x){
return (1LL << x) - 1;
}
bool bit(ll a, ll i){
return (1LL << i) & a;
}
#define llpc(x) __builtin_popcountll(x)
#define ipc(x) __builtin_popcount(x)
#define iclz(x) __builtin_clz(x)
#define llclz(x) __builtin_clzll(x)
#define ictz(x) __builtin_ctz(x)
#define llctz(x) __builtin_ctzll(x)
//answers
#define cYES cout << "YES" << endl
#define cYes cout << "Yes" << endl
#define cyes cout << "yes" << endl
#define cNO cout << "NO" << endl
#define cNo cout << "No" << endl
#define cno cout << "no" << endl
#define ipsb cout << -1 << endl
const ll mod2 = 998244353;
const ll mod = 1000000007;
const int inf = int(1e9); // ll inf = ll(1e18);
// read arr vec matr etc
#define fill(x, y) memset(x, y, sizeof(x))
void read(vector<int> &x){
for(auto &e:x) cin >> e;
}
void sread(vector<string> &x){
for(auto &e:x) cin >> e;
}
void mread(vector<vector<int>> &p, int nnn, int mmm){
for(int i = 0; i < nnn; ++i){
vector<int> pp;
for(int j = 0; j < mmm; ++j){
int wq; cin >> wq; pp.pb(wq);
}
p.pb(pp);
}
}
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); //high quality random number generator using time as seed
//int random(int l, int r){return uniform_int_distribution<int>(l,r)(rng);} //returns a randomb number between [l, r]
// Solution
vi nodes;
void tc(){
int n, k, q, t; cin >> n >> k >> q >> t;
vi a(n), b(n);
read(a); read(b);
vvi eda(n);
int root = 0;
for(int i = 0; i < n; ++i){
if(a[i] == -1){
root = i;
}else{
eda[a[i]-1].pb(i);
}
nodes.pb(i);
}
vi rootdist(n);
rootdist[root] = 0;
vi da(n);
while(sz(nodes) > k){
nodes.pop_back();
}
for(auto e:nodes){
cout << e+1 << " ";
}
cout << '\n';
vii queries;
for(int i = 1; i < k; ++i){
cout << "? " << nodes[i]+1 << " " << nodes[0]+1 << '\n';
queries.pb({i, 0});
}
cout << "!" << '\n';
cout<<flush;
fflush(stdout);
for(int i = 1; i < k; ++i){
int e, f, g, h; cin >> e >> f >> g >> h;
int u = nodes[queries[i-1].fi];
int v = nodes[queries[i-1].se];
if(da[u] < da[v]){
swap(u, v);
}
rootdist[u] = e; // calculates all distance from root on the first tree(the joint one)
//setree.join(u, v, g, h, stb); //uses graph structure to create the lca tree from the second tree, join erase a big edge, add a node, and join 2 small edges to the initial points
}
//setree.create(); //crate sum sparse table
for(int i = 0; i < t; ++i){
int u, v; cin >> u >> v;
u--;
v--;
cout << rootdist[u] + rootdist[v] - 2 * rootdist[0] << " ";
cout << rootdist[u] + rootdist[v] - 2 * rootdist[0] << " ";
cout << '\n';
}
cout<<flush;
fflush(stdout);
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int t = 1;
while(t--){
tc();
}
}
