#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 6001;
int n,m,k,s,p[MX],l[MX];
ll depth[MX];
vi child[MX];
bitset<MX> ans;
bitset<1000001> root, cyc[MX];
void dfs0(int x, ll cdepth) {
cdepth += l[x];
depth[x] = cdepth;
for (int i: child[x]) dfs0(i,cdepth);
}
void genCyc(int x, int ori, ll cdepth) {
if (x > n) return;
if (cdepth <= k) cyc[ori][cdepth] = 1;
for (int i: child[x]) genCyc(i,ori,cdepth+l[i]);
}
void DFS(int cur) {
if (cur > n) {
ll need = k-depth[cur];
if (need == 0) {
ans[cur] = 1;
return;
}
for (int t = 1; t*t <= need; t++) if (need%t == 0)
if (cyc[cur][t] || cyc[cur][need/t]) ans[cur] = 1;
}
genCyc(cur,cur,s);
for (int i: child[cur]) {
cyc[i] |= cyc[cur];
DFS(i);
}
}
void case0() {
dfs0(0,0);
DFS(0);
}
void case1() {
F0R(i,n+1) if (depth[i]+s <= k) root[depth[i]+s] = 1;
FOR(i,n+1,n+m+1) {
vi v = {l[i]};
int x = p[i];
while (x) {
v.pb(v.back()+l[x]);
x = p[x];
}
for (int x: v) if (x <= k && root[k-x]) ans[i] = 1;
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> n >> m >> k >> s; s ++;
FOR(i,1,n+1) {
cin >> p[i] >> l[i]; l[i] ++;
child[p[i]].pb(i);
}
FOR(i,n+1,n+m+1) {
cin >> p[i] >> l[i]; l[i] ++;
child[p[i]].pb(i);
}
case0();
case1();
FOR(i,n+1,n+m+1) {
if (ans[i]) cout << "YES";
else cout << "NO";
cout << "\n";
}
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
