#include <bits/stdc++.h>
#define REP(i, n)     for(int i = 0;i < n;i++)
#define REPR(i, n)    for(int i = n;i >= 0;i--)
#define FOR(i, m, n)  for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VRSORT(v) sort(v.begin(), v.end(),greater<int>());
#define ALL(X) (X).begin(),(X).end()
#define _INF 999999999999
#define M_PI 3.14159265358979323846
#define MOD 1000000007

#ifdef __LOCAL
    #define DBG(X) cout << #X << " = " << (X) << endl;
    #define SAY(X) cout << (X) << endl;
#else
    #define DBG(X)
    #define SAY(X)
#endif

#ifdef __LOCAL
    #include <filesystem>
    namespace fs = std::filesystem;
#endif

using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

int dx[]={1, 0, -1, 0, 0};
int dy[]={0, 1, 0, -1, 0};
const int INT_INF = (int)(2e9);
const ll  LL_INF = (ll)(2e18);
const int NIL = -1;
static mt19937 _g(time(nullptr));
 
std::string pad(int num){ char buffer[4]; std::snprintf(buffer, sizeof(buffer), "%03d", num); return buffer;}
inline ll randint(ll a, ll b) { ll w = (_g() << 31LL) ^ _g(); return a + w % (b - a + 1); }
inline void fast_io() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); };
template<typename T> inline T sign(T x) { return T(x > 0) - T(x < 0); }
template<typename T, typename S> inline ostream& operator<<(ostream& os, const pair<T, S> p) { cout << "[" << p.first << ";" << p.second << "]"; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const vector<T>& v) { for (auto el : v) cout << el << " "; return os; }
template<typename T> inline T fetch() { T ret; cin >> ret; return ret; }
template<typename T> inline vector<T> fetch_vec(int sz) { vector<T> ret(sz); for (auto& elem : ret) cin >> elem; return ret; }

struct edge{
    ll source,to,cost;
    edge(int a, int b, ll c):source(a),to(b),cost(c){}
};

// ワーシャルフロイド法によって計算する。
// 行列の3重ループを計算するだけで最短経路が求められる！便利！！！
// 
// なぜかいくつかのテストケースでWAになる。なんで？
int N,M;
ll L;
int Q;
vector<edge> E;
vector<int> S,T;
ll fuel[305][305],dist[305][305];

int input(){
    #ifdef __LOCAL
        fs::path p = __FILE__;
        fs::path input,output;
        input = output = p.parent_path();
        input += string("/input/") + string(p.stem()) + string(".txt");
        output += string("/output/") + string(p.stem()) + string(".txt");
        freopen(input.c_str(), "r", stdin);
        freopen(output.c_str(), "w", stdout);
    #endif
    fast_io();
    cin >> N >> M >> L;
    E.assign(M,edge(0,0,LL_INF));
    REP(i,M) {cin >> E[i].source >> E[i].to >> E[i].cost; E[i].source--; E[i].to--;}
    cin >> Q;
    S.assign(Q,0);
    T.assign(Q,0);
    REP(i,Q){
        cin >> S[i] >> T[i];
        S[i]--;
        T[i]--;
    }
    return 0;
}

int solve(){
    input();
    REP(i,N) REP(j,N) {fuel[i][j]=LL_INF; dist[i][j]=LL_INF;}
    REP(i,N) fuel[i][i] = dist[i][i] = 0LL;
    REP(i,M) fuel[E[i].source][E[i].to] = fuel[E[i].to][E[i].source] = E[i].cost; 
    REP(k,N) REP(i,N) REP(j,N) fuel[i][j] = min(fuel[i][j], fuel[i][k]+fuel[k][j]);

    REP(i,N) REP(j,N) if(fuel[i][j]<=L) dist[i][j]=1LL;
    REP(k,N) REP(i,N) REP(j,N) dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j]);

    REP(i,Q){
        if(dist[S[i]][T[i]]<LL_INF) cout << (dist[S[i]][T[i]] - 1LL) << "\n";
        else cout << "-1" << "\n";
    }
    return 0;
}

int main()
{
    solve();
    return 0;
}


Measured memory_bytes: 3584