#include <bits/stdc++.h>
using namespace std;
const int MX = 2e5 + 10;
const int MK = 45;
vector<int> adj[MX];
int N, L, Q, p[MX], h[MX], dp[MX][MK];
// dp[i][j] is the multiplication towards all of node i's descendants that are of distance d
// Update is updating all dp[anc_i][D - dist(anc_i, i)]
// But observe that this won't handle cases where we go back up and return through the path we came from
// that is why we also add dp[anc_i][D - dist(anc_i, i) - 1]
// Query is querying all dp[anc_i][dist(anc_i, i)]
void dfs(int u, int v){
p[u] = v;
for(int nxt : adj[u]){
if(nxt == v) continue;
dfs(nxt, u);
}
}
int main(){
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> N >> L;
for(int i = 1; i < N; i++){
int u, v; cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
for(int i = 1; i <= N; i++) cin >> h[i];
dfs(1, 0);
for(int i = 0; i < MX; i++)
for(int j = 0; j < MK; j++) dp[i][j] = 1;
cin >> Q;
for(int i = 1; i <= Q; i++){
int t, x, d, w; cin >> t;
if(t == 1){
cin >> x >> d >> w;
dp[x][d] = 1ll * dp[x][d] * w % L;
if(x != 1 && d > 0) dp[x][d - 1] = 1ll * dp[x][d - 1] * w % L;
int curr = x;
for(int j = 1; j <= d; j++){
if(p[curr] != 0) curr = p[curr];
dp[curr][d - j] = 1ll * dp[curr][d - j] * w % L;
if(curr != 1 && d > j) dp[curr][d - j - 1] = 1ll * dp[curr][d - j - 1] * w % L;
}
}else{
cin >> x;
int ans = h[x], curr = x;
for(int j = 0; j <= 41; j++){
// cout << "At query #" << i << " := anc " << x << " " << curr << " " << j << " " << dp[curr][j] << '\n';
ans = 1ll * ans * dp[curr][j] % L;
if(p[curr] == 0) break;
else curr = p[curr];
}
cout << ans << '\n';
}
}
}
