/*
暴力可以st表维护线性基， 从而复杂度两个log
实际上我们可以离线来做， 并且记录可行最右值， 就是一个log的了
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define mmp make_pair
#define M 300010
using namespace std;
int read() {
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
int n, m, Q, p[33], q[33], ans[M], w[M], l[M], d[M];
vector<int> note[M];
vector<pair<int, int> > to[M];
void dfs(int now, int f) {
for(int i = 0; i < to[now].size(); i++) {
int vj = to[now][i].first;
if(vj == f) continue;
d[vj] = d[now] ^ to[now][i].second;
dfs(vj, now);
}
}
int main() {
n = read(), m = read(), Q = read();
for(int i = 1; i < n; i++) {
int vi = read(), vj = read(), v = read();
to[vi].push_back(mmp(vj, v));
to[vj].push_back(mmp(vi, v));
}
dfs(1, 0);
for(int i = 1; i <= m; i++) w[i] = d[read()] ^ d[read()] ^ read();
for(int i = 1; i <= Q; i++) {
ans[i] = d[read()] ^ d[read()];
l[i] = read();
note[read()].push_back(i);
}
for(int t = 1; t <= m; t++) {
int x = w[t], r = t;
for(int i = 30; i >= 0; i--) {
if((x >> i) & 1) {
if(!p[i]) {
p[i] = x, q[i] = r;
break;
}
if(q[i] < r) swap(p[i], x), swap(q[i], r);
x ^= p[i];
}
}
for(int k = 0; k < note[t].size(); k++) {
int v = note[t][k];
for(int i = 30; i >= 0; i--) if(q[i] >= l[v]) ans[v] = min(ans[v], ans[v] ^ p[i]);
}
}
for(int i = 1; i <= Q; i++) {
cout << ans[i] << "
";
}
return 0;
}