code

//操作 1： 格式： 1 x y z 表示将树从 x 到 y 结点最短路径上所有节点的值都加上 z。
//操作 2： 格式： 2 x y表示求树从 x 到 y 结点最短路径上所有节点的值之和。
//操作 3： 格式： 3 x z表示将以 x 为根节点的子树内所有节点值都加上 z。
//操作 4： 格式： 4 x表示求以 x 为根节点的子树内所有节点值之和
#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)

int n, m, rt, tot, cnt, id[100005], top[100005], nw[100005], f[100005], dep[100005], sz[100005], mson[100005], a[100005], hd[100005], to[200005], nxt[200005];

ll mod;

struct node
{
	int l, r;
	ll sum, add;
}t[400005];

int read()
{
	int x = 0, fl = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') fl = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0'; ch = getchar();}
	return x * fl;
}

void add(int x, int y)
{
	tot ++ ;
	to[tot] = y;
	nxt[tot] = hd[x];
	hd[x] = tot;
	return;
}

void push_up(int p)
{
	t[p].sum = (t[ls(p)].sum + t[rs(p)].sum) % mod;
	return;
}

void push_down(int p)
{
	if (!t[p].add) return;
	t[ls(p)].add = (t[ls(p)].add + t[p].add) % mod;
	t[rs(p)].add = (t[rs(p)].add + t[p].add) % mod;
	t[ls(p)].sum = (t[ls(p)].sum + 1ll * t[p].add * (t[ls(p)].r - t[ls(p)].l + 1) % mod) % mod;
	t[rs(p)].sum = (t[rs(p)].sum + 1ll * t[p].add * (t[rs(p)].r - t[rs(p)].l + 1) % mod) % mod;
	t[p].add = 0;
	return;
}

void update(int p, int l0, int r0, int d)
{
	if (l0 <= t[p].l && t[p].r <= r0)
	{
		t[p].add = (t[p].add + (ll)d) % mod;
		t[p].sum = (t[p].sum + 1ll * (t[p].r - t[p].l + 1) * d % mod) % mod;
		return;
	}
	push_down(p);
	int mid = (t[p].l + t[p].r) >> 1;
	if (l0 <= mid) update(ls(p), l0, r0, d);
	if (r0 > mid) update(rs(p), l0, r0, d);
	push_up(p);
	return;
}

ll query(int p, int l0, int r0)
{
	if (l0 <= t[p].l && t[p].r <= r0) return t[p].sum;
	push_down(p);
	int mid = (t[p].l + t[p].r) >> 1; ll res = 0ll;
	if (l0 <= mid) res = (res + query(ls(p), l0, r0) % mod) % mod;
	if (r0 > mid) res = (res + query(rs(p), l0, r0) % mod) % mod;
	return res % mod;
}

void build(int p, int l0, int r0)
{
	t[p].l = l0; t[p].r = r0;
	if (l0 == r0)
	{
		t[p].sum = (ll)(nw[l0]);
		return;
	}
	int mid = (l0 + r0) >> 1;
	build(ls(p), l0, mid);
	build(rs(p), mid + 1, r0);
	push_up(p);
	return;
}

void dfs1(int x, int fa)
{
	sz[x] = 1;
	int mx = -1;
	for (int i = hd[x]; i; i = nxt[i])
	{
		int y = to[i];
		if (y == fa) continue;
		dep[y] = dep[x] + 1;
		f[y] = x;
		dfs1(y, x);
		sz[x] += sz[y];
		if (sz[y] > mx)
		{
			mx = sz[y];
			mson[x] = y;
		}
	}
	return;
}

void dfs2(int x, int tp)
{
	id[x] = ++ cnt;
	top[x] = tp;
	nw[cnt] = a[x];
	if (!mson[x]) return;
	dfs2(mson[x], tp);
	for (int i = hd[x]; i; i = nxt[i])
	{
		int y = to[i];
		if (y == f[x] || y == mson[x]) continue;
		dfs2(y, y);
	}
	return;
}

void q1()
{
	int x = read(), y = read(), z = read();
	while (top[x] != top[y])
	{
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		update(1, id[top[x]], id[x], (ll)z);
		x = f[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	update(1, id[x], id[y], (ll)z);
	return;
}

void q2()
{
	int x = read(), y = read();
	ll res = 0ll;
	while (top[x] != top[y])
	{
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		res = (res + query(1, id[top[x]], id[x])) % mod;
		x = f[top[x]];
	}
	if (dep[x] > dep[y]) swap(x, y);
	res = (res + query(1, id[x], id[y])) % mod;
	printf("%lld\n", res);
	return;
}

void q3()
{
	int x = read(), z = read();
	update(1, id[x], id[x] + sz[x] - 1, (ll)z);
	return;
}

void q4()
{
	int x = read();
	printf("%lld\n", query(1, id[x], id[x] + sz[x] - 1) % mod);
	return;
}

int main()
{
	n = read(); m = read(); rt = read(); mod = (ll)read();
	for (int i = 1; i <= n; i ++ ) a[i] = read();
	for (int i = 1; i <= n - 1; i ++ )
	{
		int x = read(), y = read();
		add(x, y); add(y, x);
	}
	dfs1(rt, 0); dfs2(rt, rt);
	build(1, 1, n);
	while (m -- )
	{
		int opt = read();
		if (opt == 1) q1();
		else if (opt == 2) q2();
		else if (opt == 3) q3();
		else q4();
	}
	return 0;
}