【题目链接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=3545
【题解】
　　显然离线后会好做许多。
　　将所有边与询问的困难值排序，用并查集维护连通性，用权值线段树维护连通块权值。每次加入一条边时，若在不同连通块中，就将这两个块合并。用线段树合并维护权值线段树。查询时查询所在块的第K大即可。
　　时间复杂度$O(N*logN)$
　　此题也有在线做法，见：[bzoj3551]
【代码】

/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :   [bzoj3545]
    Points :    segment trees' merge
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    N       500010
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
const int L = 0, R = 1e9;
struct Edge{
    int u, v, w, id;
}e[N], p[N];
struct Tree{
    int pl, pr, num;
}T[N * 18];
int f[N], place, n, m, q, h[N], rt[N], ans[N];
bool cmp(Edge x, Edge y){
    return x.w < y.w;
}
int dad(int x){
    if (f[x] == x) return f[x];
        return f[x] = dad(f[x]);
}
int extend(int p, int data, int l, int r){
    if (!p) p = ++place;
    T[p].num++;
    if (l != r){
        int mid = (l + r) / 2;
        if (mid >= data) T[p].pl = extend(T[p].pl, data, l, mid);
            else T[p].pr = extend(T[p].pr, data, mid + 1, r);
    }
    return p;
} 
int merge(int p1, int p2, int l, int r){
    if (p1 == 0) return p2;
    if (p2 == 0) return p1;
    T[p1].num = T[p1].num + T[p2].num;
    if (l != r){
        int mid = (l + r) / 2;
        T[p1].pl = merge(T[p1].pl, T[p2].pl, l, mid);
        T[p1].pr = merge(T[p1].pr, T[p2].pr, mid + 1, r);
    }
    return p1;
}
int query(int p, int data, int l, int r){
    if (l == r) return l;
    int mid = (l + r) / 2;
    if (T[T[p].pr].num >= data)
        return query(T[p].pr, data, mid + 1, r);
        else return query(T[p].pl, data - T[T[p].pr].num, l, mid);
}
int main(){
//  freopen(".in", "r", stdin);
//  freopen(".out", "w", stdout);
    n = read(), m = read(), q = read();
    for (int i = 1; i <= n; i++){
        h[i] = read(); f[i] = i;
        rt[i] = extend(rt[i], h[i], L, R);
    }
    for (int i = 1; i <= m; i++)
        e[i].u = read(), e[i].v = read(), e[i].w = read();
    sort(e + 1, e + m + 1, cmp);
    for (int i = 1; i <= q; i++)
        p[i].u = read(), p[i].w = read(), p[i].v = read(), p[i].id = i;
    sort(p + 1, p + q + 1, cmp);
    for (int i = 1, el = 1, pl = 1; i <= m + q; i++){
        if (pl > q || (el <= m && e[el].w <= p[pl].w)){
            int u = dad(e[el].u), v = dad(e[el].v);
            if (u != v){
                rt[u] = merge(rt[u], rt[v], L, R);
                f[v] = u;
            }
            el++;
        }
        else {
            int u = dad(p[pl].u);
            if (T[rt[u]].num < p[pl].v)
                ans[p[pl].id] = -1; 
                else ans[p[pl].id] = query(rt[u], p[pl].v, L, R);
            pl++;
        } 
    }
    for (int i = 1; i <= q; i++)
        printf("%d\n", ans[i]);
    return 0;
}