/*

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] << "\n";

	}

	return 0;

}