#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define eps 1e-9
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dep(i,a,b) for(int i=(a);i>=(b);i--)
#define lowbit(a) (a)&(-a)
using namespace std;
const int maxn = 600005; // 点数N加上虚拟节点N-1，最多 2N 个点，约 60万
const int mo = 998244353;
const double pi = acos(-1.0);
template <typename T>
inline void read(T &X)
{
    X = 0;int w = 0; char ch = 0;
    while(!isdigit(ch)) {w |= ch == '-' ;ch = getchar();}
    while(isdigit(ch)) X = X * 10 + (ch^48),ch = getchar();
    if(w) X = -X;
}
int n,m,q;
struct Edge
{
    int u,v,w;
}e[maxn];
bool cmp(Edge A,Edge B)
{
    return A.w < B.w;
}
int fa[maxn];
int find(int u)
{
    if(u == fa[u]) return u;
    return fa[u] = find(fa[u]);
}
int val[maxn];
int ch[maxn][2];
int dep[maxn];
int f[maxn][20];
void dfs(int u,int father)
{
    dep[u] = dep[father] + 1;
    f[u][0] = father;
    rep(i,1,19) f[u][i] = f[f[u][i-1]][i-1];
    if(ch[u][0]) dfs(ch[u][0],u);
    if(ch[u][1]) dfs(ch[u][1],u);
}
int lca(int x,int y)
{
    if(dep[x] < dep[y]) swap(x,y);
    dep(i,19,0)
    {
        if(dep[f[x][i]] >= dep[y]) x = f[x][i];
    }
    if(x == y) return x;
    dep(i,19,0)
    {
        if(f[x][i] != f[y][i])
        {
            x = f[x][i];
            y = f[y][i];
        }
    }
    return f[x][0];
}
void solve()
{
    read(n),read(m),read(q);
    rep(i,1,m) read(e[i].u),read(e[i].v),read(e[i].w);
    sort(e+1,e+1+m,cmp);
    rep(i,1,n*2) fa[i] = i; 
    int tot = n; 
    rep(i,1,m)
    {
        int fu = find(e[i].u);
        int fv = find(e[i].v);
        if(fu != fv)
        {
            tot++;
            val[tot] = e[i].w; // 虚拟节点的点权即为边权
            ch[tot][0] = fu;
            ch[tot][1] = fv;
            fa[fu] = tot;
            fa[fv] = tot;
        }
    }
    rep(i,1,tot)
    {
        if(fa[i] == i) dfs(i,0); // 找到每棵树的根节点，向下预处理倍增数组
    }
    while(q--)
    {
        int u,v;
        read(u),read(v);
        if(find(u) != find(v)) cout << -1 << '\n'; // 30万次询问，为防止TLE，此处作特殊情况处理，使用\n替换endl
        else cout << val[lca(u,v)] << '\n';
    }
    return;
}
int main()
{
    int T=1;
//  freopen("mul.in","r",stdin);
//  freopen("mul.out","w",stdout);
//  read(T);
    while(T--) solve();
    return 0;
}