#include <bits/stdc++.h>
#define fi first
#define se second
#define For(i, a, b) for (int i = (a); i <= (b); ++i)
#define ForD(i, a, b) for (int i = (a); i >= (b); --i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repD(i, n) for (int i = (n) - 1; i >= 0; --i)
#define coutE(x) {cout << x << '\n'; return;}
#define pb push_back
#define pf push_front
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define bs binary_search
#define ub upper_bound
#define lb lower_bound
#define endl '\n'
#define db long double
#define ll long long
#define ii pair<int, int>
#define vi vector<int>
#define vii vector<ii>
using namespace std;
const int N = 1e5 + 5;
const int B = 1000;
int n, m, q;
struct edge{
int u, v, w;
} ed[N];
struct query{
int t, x, w;
} quer[N];
bool bad[N];
int ans[N];
vi g[N];
int par[50005];
stack<pair<int&, int>> history;
int find(int u){
while (par[u] > 0) u = par[u];
return u;
}
void rollBack(int t){
while (sz(history) > t){
history.top().fi = history.top().se;
history.pop();
}
}
void Union(edge e){
int u = find(e.u), v = find(e.v);
if (u == v) return;
if (par[u] > par[v]) swap(u, v);
history.push({par[u], par[u]});
history.push({par[v], par[v]});
par[u] += par[v];
par[v] = u;
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n >> m;
For(i, 1, m){
cin >> ed[i].u >> ed[i].v >> ed[i].w;
}
cin >> q;
For(i, 1, q){
cin >> quer[i].t >> quer[i].x >> quer[i].w;
}
for (int l = 1; l <= q; l += B){
memset(par, -1, sizeof par);
while (!history.empty()) history.pop();
int r = min(q, l + B - 1);
vi ask, changed, unchanged;
For(i, l, r){
if (quer[i].t == 1){
//change is bad
bad[quer[i].x] = 1;
//remember that edge's id not query's id
}
else{
ask.pb(i);
}
}
For(i, 1, m){
if (bad[i])
changed.pb(i);
else
unchanged.pb(i);
//remember to reset
bad[i] = 0;
}
//O(B^2)
For(i, l, r){
if (quer[i].t == 1){
//update new weight
ed[quer[i].x].w = quer[i].w;
}
else{
for (int j: changed){
if (ed[j].w >= quer[i].w){
//g[i] saves bad edges that i needs
g[i].pb(j);
}
}
}
}
sort(all(ask), [&](int i, int j){
return quer[i].w > quer[j].w;
});
sort(all(unchanged), [&](int i, int j){
return ed[i].w > ed[j].w;
});
int pnt = 0;
//O(B * (m + B))
for (int i: ask){
while (pnt < sz(unchanged) && ed[unchanged[pnt]].w >= quer[i].w)
Union(ed[unchanged[pnt++]]);
int last = sz(history);
for (int j: g[i]) Union(ed[j]);
ans[i] = -par[find(quer[i].x)];
rollBack(last); //O(B)
//reset
g[i].clear();
}
}
For(i, 1, q) if (quer[i].t == 2)
cout << ans[i] << endl;
return 0;
}
