【P3384】 树链剖分

有四个操作 $1$ 操作是$x-y$都加上 $z$
$2$ 操作是求 $x-y$的点值和
$3$ 操作是使x节点所有子树+ z
$4$ 操作是求x节点所有子树和
我们对1操作来说 利用重链去跳 可以减少复杂度
对2操作来说 利用重链去查询
对3操作来说 直接查子树 id[x]+sz[x]-1正好是x子树 (dfs序性质
然后4操作 也是这个原理
我们只要耐心敲一下 一定会$De$出 $BUG$的 $滑稽$

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

const int MAX_N = 100025;
int n,mod,cnt;
int p[MAX_N],eid,fa[MAX_N][20],lg[MAX_N],sz[MAX_N],depth[MAX_N],son[MAX_N],Rank[MAX_N],top[MAX_N],id[MAX_N],arr[MAX_N];
ll s[MAX_N<<2],col[MAX_N<<2];
void up(int rt)
{
    s[rt] = (s[rt<<1] + s[rt<<1|1])%mod;
}
void down(int rt,int l,int r)
{
    if(col[rt])
    {
        int mid = (l+r)>>1;
        col[rt<<1] += col[rt];
        col[rt<<1|1] += col[rt];
        s[rt<<1] += (mid-l+1)*col[rt];
        s[rt<<1|1] += (r-mid)*col[rt];
        s[rt<<1] %=mod;
        s[rt<<1|1] %= mod;
        col[rt] = 0;
    }
}
void build(int rt,int l,int r)
{
    if(l==r)
    {
        s[rt] = arr[Rank[l]]%mod;
        return ;
    }
    int mid = (l+r)>>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
    up(rt);
}
void update(int rt,int l,int r,int x,int y,int v)
{
    if(x<=l&&r<=y)
    {
        col[rt] += v;
        s[rt] += (r-l+1)*v;
        return ;
    }
    int mid =(l+r)>>1;
    down(rt,l,r);
    if(x<=mid) update(rt<<1,l,mid,x,y,v);
    if(mid<y) update(rt<<1|1,mid+1,r,x,y,v);
    up(rt);
}
ll query(int rt,int l,int r,int x,int y)
{
    if(x<=l&&r<=y)
    {
        return s[rt];
    }
    int mid = (l+r)>>1;
    ll res = 0;
    down(rt,l,r);
    if(x<=mid) res+=query(rt<<1,l,mid,x,y);
    if(mid<y) res+= query(rt<<1|1,mid+1,r,x,y);
    return (res%mod);
}
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
struct edge
{
    int v,next;
}e[MAX_N<<1];
void add(int u,int v)
{
    e[eid].v = v;
    e[eid].next = p[u];
    p[u] = eid++;
}
void dfs1(int x,int fath)
{
//    dbg(x);
    sz[x] = 1;
    depth[x] = depth[fath] + 1;
    fa[x][0] = fath;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int to = e[i].v;
//        dbg2(to,fath);
        if(to==fath) continue;
        dfs1(to,x);
        sz[x]+=sz[to];
        if(sz[to]>sz[son[x]])
            son[x] = to;
    }
}
void dfs2(int u,int t)
{
    top[u] = t;
    id[u] = ++cnt;
    Rank[cnt] = u;
    if(!son[u])
        return ;
    dfs2(son[u],t);
    for(int i = p[u];i+1;i=e[i].next)
    {
        int to = e[i].v;
        if(to!=son[u]&&to!=fa[u][0])
            dfs2(to,to);
    }
}
ll sum(int x,int y)
{
    ll ans = 0;
    while(top[x]!=top[y])
    {
        if(depth[top[x]]<depth[top[y]])
            swap(x,y);
        (ans+=query(1,1,n,id[top[x]],id[x]))%=mod;
        x = fa[top[x]][0];
    }
    if(id[x]>id[y])
        swap(x,y);
    return (ans+query(1,1,n,id[x],id[y]))%mod;
}
void updates(int x,int y,int c)
{
    while(top[x]!=top[y])
    {
        if(depth[top[x]]<depth[top[y]])
            swap(x,y);
        update(1,1,n,id[top[x]],id[x],c);
        x = fa[top[x]][0];
    }
    if(id[x]>id[y])
        swap(x,y);
    update(1,1,n,id[x],id[y],c);
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int Q,root,opt,x,y,z,a,b;
    init();
    scanf("%d%d%d%d",&n,&Q,&root,&mod);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d",&a,&b);
        add(a,b);add(b,a);
    }
    dfs1(root,0);
    dfs2(root,root);
    build(1,1,n);
    while(Q--)
    {
        scanf("%d",&opt);
        if(opt==1)
        {
            scanf("%d%d%d",&x,&y,&z);
            updates(x,y,z);
        }
        else if(opt==2)
        {
            scanf("%d%d",&x,&y);
            printf("%d\n",sum(x,y));
        }
        else if(opt==3)
        {
            scanf("%d%d",&x,&z);
            update(1,1,n,id[x],id[x]+sz[x]-1,z);
        }
        else
        {
            scanf("%d",&x);
            printf("%d\n",query(1,1,n,id[x],id[x]+sz[x]-1));
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}