评价一下，周六和周一调的半死，今天看一眼就知道哪里错了。
只经过 $\leq x$ 的路径，容易想到 Kruskal 重构树，然后看到第 $k$ 大，容易想到主席树。
那就做完了。3.43kb 代码也不过如此。
下面放的是有强制在线的代码。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 15, M = 1e6 + 15;
int n, n1, cnt, m, q, a[N], b[N];

int rt[N];
struct Segment_Tree {
    int tot = 0;
    struct Tree {
        int ls, rs;
        int sum;
    } tr[N << 5];
    void build(int &u, int l, int r) {
        u = ++tot, tr[u].sum = 0;
        if (l == r) return;
        int mid = l + r >> 1;
        build(tr[u].ls, l, mid);
        build(tr[u].rs, mid + 1, r);
    }
    void change(int &u, int l, int r, int x) {
        tr[++tot] = tr[u], u = tot;
        if (l == r) { tr[u].sum++; return; }
        int mid = l + r >> 1;
        if (x <= mid) change(tr[u].ls, l, mid, x);
        else change(tr[u].rs, mid + 1, r, x);
        tr[u].sum = tr[tr[u].ls].sum + tr[tr[u].rs].sum;
    }
    int query(int u, int v, int l, int r, int k) { //线段树二分
        if (l == r) return l;
        int mid = l + r >> 1, delta = tr[ tr[u].rs ].sum - tr[ tr[v].rs ].sum;
        if (k <= delta) return query(tr[u].rs, tr[v].rs, mid + 1, r, k);
        else return query(tr[u].ls, tr[v].ls, l, mid, k - delta);
    }
    int qry(int l, int r, int k) { return query(rt[r], rt[l - 1], 1, cnt, k); }
} t;

struct edges {
    int u, v, w;
} edge[M];
bool cmp(edges a, edges b) { return a.w < b.w; }
int p[N];
int find(int x) { return (x == p[x]) ? x : p[x] = find(p[x]); }

int h[N], e[M], ne[M], idx = 0;
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }

int l[N], r[N], w[N], high[N], tot = 0;
//w: 虚点代表的边权；  high：原图中山的高度（离散化后）
int fa[N][25], dep[N];

void dfs(int u, int father) {
    fa[u][0] = father, dep[u] = dep[father] + 1;
    if (u <= n) {
        ++tot;
        l[u] = r[u] = tot, high[tot] = a[u];
        return;
    }
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (v == father) continue;
        dfs(v, u);
        l[u] = min(l[u], l[v]);
        r[u] = max(r[u], r[v]);
    }
}
void init() {
    for (int i = 1; i <= 20; i++)
        for (int j = 1; j <= n1; j++) fa[j][i] = fa[fa[j][i - 1]][i - 1];
}

int main() {
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    sort(b + 1, b + 1 + n);
    cnt = unique(b + 1, b + 1 + n) - b - 1;
    for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + cnt, a[i]) - b;

    for (int i = 1; i <= m; i++) scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);
    sort(edge + 1, edge + 1 + m, cmp);

    for (int i = 1; i <= 2 * n; i++) p[i] = i, h[i] = -1, l[i] = 0x3f3f3f3f, r[i] = 0;
    n1 = n;
    for (int i = 1; i <= m; i++) {
        int u = edge[i].u, v = edge[i].v;
        int fu = find(u), fv = find(v);
        if (fu == fv) continue;
        n1++, p[fu] = n1, p[fv] = n1;
        w[n1] = edge[i].w;
        add(n1, fu), add(fu, n1);
        add(n1, fv), add(fv, n1);
        // cout << '\t' << n1 << ' ' << fu << ' ' << w[n1] << endl;
        // cout << '\t' << n1 << ' ' << fv << ' ' << w[n1] << endl;
    }
    dfs(n1, 0);
    init();
    // for (int i = 1; i <= n1; i++) cout << i << ' ' << l[i] << ' ' << r[i] << endl;
    t.build(rt[0], 1, cnt);
    for (int i = 1; i <= tot; i++) rt[i] = rt[i - 1], t.change(rt[i], 1, cnt, high[i]);
    int lst = 0;
    while (q--) {
        int x, lim, k; scanf("%d%d%d", &x, &lim, &k);
        x = (x ^ lst) % n + 1, k = (k ^ lst) % n + 1, lim ^= lst;
        for (int i = 20; i >= 0; i--)
            if (fa[x][i] && w[fa[x][i]] <= lim) x = fa[x][i];
        // cout << '\t' << x << endl;
        if (r[x] - l[x] + 1 < k) { lst = 0; puts("-1"); continue; }
        printf("%d\n", lst = b[t.qry(l[x], r[x], k)]);
    }
    return 0;
}