• 這道題我們首先看到lca會想到倍增，但是發現還有合併的操作
• 這就尷尬了，考場上直接打了個暴力然後愉快的報零了
• 這時我們也不慌，下來之後我們也沒想到好方法，看網上題解發現居然用LCT（我沒學）
• 當然博主不是一個想學新方法的人
• 於是開始鑽研線上操作
• 我們對於每次操作直接將原來的圖重建一次
• 然後強行再次運算倍增
• 為了避免超時當然要加上輸入輸出優化
• 下面是程式碼：

#include<bits/stdc++.h>
using namespace std;
const int Max = 1e5+5;
int fa[Max][21],d[Max],
	  rt[Max],root[Max]
 
,
	  size[Max],head[Max * 2],
	  cnt = 0;
void read(int &x)
{
	int f = 1;
	x = 0;
	char c = getchar();
	while (c < '0' || c > '9')
	{
		if (c == '-') f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
	{
		x = x * 10 + c - '0';
		c = getchar();
	}
	x *= f;
	return;
}
void write
 
(int x)
{
	if (x < 0) putchar('-'),x = -x;
	if (x > 9) write(x/10);
	putchar(x % 10 + '0');
}
struct node
{
	int next,to;
}e[Max * 2];
void add(int u,int v)
{
	e[++cnt].next = head[u];
	e[cnt].to = v;
	head[u] = cnt;
	e[++cnt].next = head[v];
	e[cnt].to = u;
	head[v] = cnt;
}
int n,m,q,E;
void rebuild
 
(int x,int fr)
{
	for (int i = 1; i <= 20; i++)
		fa[x][i] = fa[fa[x][i-1]][i-1];
	for (int i = head[x]; i; i = e[i].next)
	{
		int to = e[i].to;
		if (to == fr) continue;
		d[to] = d[x] + 1;
		fa[to][0] = x;
		rebuild(to,x);
	}
}
void dfs(int x, int fr)
{
	fa[x][0] = fr;
	for (int i = head[x]; i; i = e[i].next)
	{
		int to = e[i].to;
		if (to == fr) continue;
		d[to] = d[x] + 1;
		dfs(to,x);
		size[x] += size[to];
	}
}
int find(int x)
{
	for (int i = 20; i >= 0; i--)
		if(fa[x][i]) x = fa[x][i];
	return x;
}
int lca(int x,int y)
{
	if (d[x] < d[y]) swap(x,y);
	for (int i = 20; i >= 0; i--)
		if(d[x] - (1 << i) >= d[y])
			x = fa[x][i];
	if (x == y) return x;
	for (int i = 20; i >= 0; i--)
		if(fa[x][i] != fa[y][i])
		{
			x = fa[x][i];
			y = fa[y][i];
		}
	return fa[x][0];
}
int main()
{
//	freopen("data.in","r",stdin);
//	freopen("data.out","w",stdout);
	read(n);
	read(m);
	for (int i = 1; i <= m; i++)
		read(rt[i]);
	read(E);
	
	
	for (int i = 1; i <= n-m; i++)
	{
		int x,y;
		read(x);
		read(y);
		add(x,y);
	}
	
	
	for (int i = 1; i <= n; i++)
		size[i] = 1;
	for (int i = 1; i <= m; i++)
	{
		d[rt[i]] = 1;
		dfs(rt[i],0);
	}
	for (int i = 1; i <= 20; i++)
		for (int j = 1; j <= n; j++)
			fa[j][i] = fa[fa[j][i - 1]][i - 1];
	for (int i = 1; i <= n; i++)
	{
		int o = find(i);
		if (o) root[i] = o;
		else root[i] = i;
	}
	
	
	read(q);
	while (q--)
	{
		int op,u,v;
		read(op);
		read(u);
		read(v);
		if (op == 1)
		{
			int x = find(u),
				  y = find(v);
			if (x == y) continue;
			if (size[x] > size[y])
			{
				fa[v][0] = u;
				size[x] += size[y];
				d[v] = d[u] + 1;
				root[y] = root[x];
				add(u,v);
				rebuild(v,u);
			}
			else
			{
				fa[u][0] = v;
				size[y] += size[x];
				d[u] = d[v] + 1;
				root[y] = root[x];
				add(u,v);
				rebuild(u,v);
			}
		}
		else
		{
			if (find(u) != find(v))
			{
				printf("orzorz\n");
				continue;
			}
			int x = find(u);
			int o = lca(u,v),
				  t = lca(root[x],v);
			if (d[o] < d[t]) o = t;
			t = lca(root[x],u);
			if (d[o] < d[t]) o = t;
			write(o);
			putchar('\n');
		}
	}
	return 0;
}