#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++ i);
#define per(i, a, n) for(int i = n; i >= a; -- i);
typedef long long ll;
const int N = 2e6+ 5;
// const ll mod = 1e9 + 7;
int mod;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;
const int G = 3, Gi = 332748118;
ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b);}
bool cmp(int a, int b){ return a > b;}
//
int n, m, r;
int head[N], cnt = 0;
struct node{
int to, nxt, c;
}edge[N << 1];
struct Tree{
int l, r, val, lz;
}tree[N * 4];
int val[N], tval[N];
int son[N], siz[N],dfn[N], dep[N], top[N], fa[N], rnk[N];
int res = 0, tot = 0;
void add(int u, int v){
edge[cnt].to = v, edge[cnt].nxt = head[u], head[u] = cnt ++;
edge[cnt].to = u, edge[cnt].nxt = head[v], head[v] = cnt ++;
}
void pushdown(int index){
if(tree[index].lz){
tree[index << 1].val += (tree[index << 1].r - tree[index << 1].l + 1) * tree[index].lz % mod;
tree[index << 1 | 1].val += (tree[index<<1|1].r - tree[index<<1|1].l + 1) * tree[index].lz % mod;
tree[index << 1].lz += tree[index].lz;
tree[index << 1 | 1].lz += tree[index].lz;
tree[index].lz = 0;
}
}
void pushup(int index){
tree[index].val = (tree[index << 1].val + tree[index << 1 | 1].val) % mod;
}
void Build(int l, int r, int index){
tree[index].l = l, tree[index].r = r;
tree[index].lz = 0;
if(l == r){
tree[index].val = tval[l] % mod;
return;
}
int mid = (l + r) >> 1;
Build(l, mid, index << 1);
Build(mid + 1, r, index << 1 | 1);
pushup(index);
}
void updata(int l, int r, int index, int val){
if(tree[index].l >= l && tree[index].r <= r){
tree[index].lz += val;
tree[index].val += (tree[index].r - tree[index].l + 1) * val;
tree[index].val %= mod;
return;
}
if(tree[index].lz) pushdown(index);
int mid = (tree[index].l + tree[index].r) >> 1;
if(l <= mid) updata(l, r, index << 1, val);
if(r > mid) updata(l, r, index << 1 | 1, val);
pushup(index);
}
int query(int l, int r, int index){
if(l <= tree[index].l && tree[index].r <= r){
return tree[index].val % mod;
}
if(tree[index].lz) pushdown(index);
int mid = (tree[index].l + tree[index].r) >> 1;
int ans = 0;
if(l <= mid) ans += query(l, r, index << 1);
if(r > mid) ans += query(l, r, index << 1 | 1);
return ans % mod;
}
// --------------------------------
int qRange(int x, int y){ //x 到 y树上最短路径结点权值和
int res = 0;
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
res += query(dfn[top[x]], dfn[x], 1);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
res += query(dfn[x], dfn[y], 1);
return res % mod;
}
void updRange(int x, int y, int c){ //x 到 y最短路径上点值 + z
c %= mod;
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
updata(dfn[top[x]], dfn[x], 1, c);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
updata(dfn[x], dfn[y], 1, c);
}
int qSon(int x){ //以x为根结点的子树内所有节点值之和
return query(dfn[x], dfn[x] + siz[x] - 1, 1);
}
void updSon(int x, int val){ //以x为根的子树内所有节点值 + z
updata(dfn[x], dfn[x] + siz[x] - 1, 1, val);
}
void dfs1(int u, int pre){
dep[u] = dep[pre] + 1;
fa[u] = pre;
siz[u] = 1;
int maxx = -1;
for(int i = head[u]; i != -1; i = edge[i].nxt){
int v = edge[i].to;
if(v == pre) continue;
dfs1(v, u);
siz[u] += siz[v];
if(siz[v] > maxx){
maxx = siz[v];
son[u] = v;
}
}
}
void dfs2(int u, int topu){ //topu当前链的最顶端的节点
dfn[u] = ++ tot;
tval[tot] = val[u];
top[u] = topu;
rnk[tot] = u;
if(!son[u]) return;
dfs2(son[u], topu);
for(int i = head[u]; i != -1; i = edge[i].nxt){
int v = edge[i].to;
if(v == son[u] || v == fa[u]) continue;
dfs2(v, v);
}
}
int main()
{
scanf("%d%d%d%d",&n, &m, &r, &mod);
cnt = 0; head[0] = -1;
for(int i = 1; i <= n; ++ i) {
scanf("%d",&val[i]);
head[i] = -1;
}
for(int i = 1; i < n; ++ i){
int x, y; scanf("%d%d",&x,&y);
add(x, y);
}
dfs1(r, r);
dfs2(r, r);
Build(1, n, 1);
while(m --){
int k, x, y, z;
scanf("%d",&k);
if(k == 1){ //x 到 y最短路径上点值 + z
scanf("%d%d%d",&x,&y,&z);
updRange(x, y, z);
}
else if(k == 2){ //x 到 y树上最短路径结点权值和
scanf("%d%d",&x,&y);
printf("%d
",qRange(x, y));
}
else if(k == 3){ //以x为根的子树内所有节点值 + z
scanf("%d%d",&x,&y);
updSon(x, y);
}
else{ //以x为根结点的子树内所有节点值之和
scanf("%d",&x);
printf("%d
",qSon(x));
}
}
return 0;
}