#include<bits/stdc++.h>
using namespace std;
int n,m;
int dep[200005],lca[200005][19],dfn[200005],ssz[200005];
vector<int>p[200005];
int trie[40000005][2];long long sz[40000005];int nw = 0;
struct node{
	int l,r,rt;
}tree[800005];
inline void ins(int now,int x,int k){
	sz[now]+=k;
	for(int i = 29;i>=0;i--){
		bool a = x>>i&1;
		if(!trie[now][a])trie[now][a] = ++nw;
		now = trie[now][a];sz[now]+=k;
	}
}inline long long query(int now,int a,int b){
	long long ans = 0;
	//cout << "  " << now << " " << a << " " << b << endl;
	for(int i = 29;i>=0;i--){if(!now)break;
		bool aa = a>>i&1,bb = b>>i&1;
	//	cout << " " << aa << " " << bb << endl;
		if(bb == 0){
			now = trie[now][aa];
		}else{
		//	cout << "   " << trie[now][a] << " " << sz[trie[now][aa]] << endl;
			ans+=sz[trie[now][aa]];now = trie[now][1^aa];
		}
	}return ans+sz[now];
	//x^a<b
}
void build(int l,int r,int k){
	tree[k].l = l,tree[k].r = r;
	tree[k].rt = ++nw;
	if(l+1 == r){
		return;
	}build(l,l+r>>1,k<<1),build(l+r>>1,r,k<<1|1);
}void update(int l,int r,int x,int nn,int k){
	int ll = tree[k].l,rr = tree[k].r;
	if(ll>=r or l>=rr)return;
	if(l<=ll and rr<=r){
		ins(tree[k].rt,x,nn);return;
	}update(l,r,x,nn,k<<1),update(l,r,x,nn,k<<1|1);
}long long qquery(int pos,int k,int a,int b){
	int l = tree[k].l,r = tree[k].r;
	long long rr = query(tree[k].rt,a,b);
	if(l+1 == r)return rr;
	int mid =l +r>>1;
	if(pos<mid)return qquery(pos,k<<1,a,b)+rr;
	return qquery(pos,k<<1|1,a,b)+rr;
}
int cc;
void dfs(int now,int fa){
	ssz[now] = 1,dep[now] = dep[fa]+1,dfn[now] = ++cc;
	lca[now][0] = fa;for(int i = 1;i<=18;i++)lca[now][i] = lca[lca[now][i-1]][i-1];
	for(int i= 0;i<p[now].size();i++){
		if(p[now][i] == fa)continue;
		dfs(p[now][i],now);ssz[now]+=ssz[p[now][i]];
	}
}int kth(int x,int k){
	for(int i =0;k;k>>=1,i++)if(k&1)x = lca[x][i];return x;
}
int lcaa(int x,int y){
	if(dep[x]<dep[y])swap(x,y);
	int dis = dep[x]-dep[y];
	for(int i =0;dis;dis>>=1,i++)if(dis&1)x = lca[x][i];
	if(x == y)return x;
	for(int i = 18;i>=0;i--){
		if(lca[x][i]!=lca[y][i]){
			x = lca[x][i],y = lca[y][i];
		}
	}return lca[x][0];
}int nrt = 1;
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> n;
	for(int i = 1;i<n;i++){
		int a,b;cin >> a >> b;
		p[a].push_back(b),p[b].push_back(a);
	}dfs(1,1);build(1,n+1,1);
	int q;cin >> 	q;
	while(q--){
		int op;cin >> op;
		if(op == 1){
			//假如在根和1之间需要特判
			int v,x,k;cin >> v >> x >> k;
			if(lcaa(nrt,v) == v){
				update(1,n+1,x,k,1);
				if(v == nrt){
					continue;
				}
				int nxt = kth(nrt,abs(dep[nrt]-dep[v])-1);
				update(dfn[nxt],dfn[nxt]+ssz[nxt],x,-k,1);
			}else{
				update(dfn[v],dfn[v]+ssz[v],x,k,1);
			}
		}if(op == 2){
			int v,a,b;cin >> v >> a >> b;
			long long ans = qquery(dfn[v],1,a,b);
			cout << ans << '\n';
		}if(op == 3){
			cin >> nrt;
		}
	}
	return 0;
}/*
这么大的 DS 题你不给我大洋里我写牛魔呢 /wx/wx/wx
线段树套 trie，换根很套路。

4
1 2
1 3
3 4
14
1 1 1 1
1 2 2 1
1 3 3 1
2 1 0 4
2 2 1 3
2 4 3 2
3 4
1 3 3 1
1 4 4 1
1 1 1 1
1 2 2 1
2 1 0 4
2 2 1 3
2 4 3 2

1
2
1 1 3 2
2 1 1 2
*/