Firstly, we use dfs order to transform the subtree into intervals. We need to maintain the interval addition and the number of different prime numbers of interval% m.
Because m is only 1000, we can maintain a bitset of 1000 size for each node that is directly violent.
One bit of bitset is only 1 bit, so the memory is 1000/8 bytes. So the spatial complexity is O(4n 1000/8)
The bitset operations are O(1000/32), so our overall complexity is O(q(logn1000/32+tot)).
Among them, tot is a prime number, not more than 200.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <bitset>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,m,a[N],h[N],num=0,id[N],in[N],out[N],dfn=0,prime[200],tot=0;
bool notprime[1000];
struct node{
    bitset<1000>x;int add;
    friend node operator+(node a,node b){
        a.x|=b.x;a.add=0;return a;
    }
}tr[N<<2];
struct edge{
    int to,next;
}data[N<<1];
inline void getprime(){
    notprime[1]=1;
    for(int i=2;i<m;++i){
        if(!notprime[i]) prime[++tot]=i;
        for(int j=1;prime[j]*i<m;++j){
            notprime[prime[j]*i]=1;
            if(i%prime[j]==0) break;
        }
    }
}
void dfs(int x,int Fa){
    in[x]=++dfn;id[dfn]=x;
    for(int i=h[x];i;i=data[i].next){
        int y=data[i].to;if(y==Fa) continue;dfs(y,x);
    }out[x]=dfn;
}
inline void build(int p,int l,int r){
    if(l==r){tr[p].x[a[id[l]]]=1;return;}
    int mid=l+r>>1;build(p<<1,l,mid);build(p<<1|1,mid+1,r);
    tr[p]=tr[p<<1]+tr[p<<1|1];
}
inline void doadd(int p,int val){
    tr[p].x=(tr[p].x<<val)|(tr[p].x>>m-val);tr[p].add+=val;tr[p].add%=m;
}
inline void pushdown(int p){
    if(!tr[p].add) return;
    doadd(p<<1,tr[p].add);doadd(p<<1|1,tr[p].add);tr[p].add=0;
}
inline void add(int p,int l,int r,int x,int y,int val){
    if(x<=l&&r<=y){doadd(p,val);return;}
    int mid=l+r>>1;pushdown(p);
    if(x<=mid) add(p<<1,l,mid,x,y,val);
    if(y>mid) add(p<<1|1,mid+1,r,x,y,val);
    tr[p]=tr[p<<1]+tr[p<<1|1];
}
inline node ask(int p,int l,int r,int x,int y){
    if(x<=l&&r<=y) return tr[p];
    int mid=l+r>>1;pushdown(p);
    if(y<=mid) return ask(p<<1,l,mid,x,y);
    if(x>mid) return ask(p<<1|1,mid+1,r,x,y);
    return ask(p<<1,l,mid,x,y)+ask(p<<1|1,mid+1,r,x,y);
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();m=read();getprime();
    for(int i=1;i<=n;++i) a[i]=read()%m;
    for(int i=1;i<n;++i){
        int x=read(),y=read();
        data[++num].to=y;data[num].next=h[x];h[x]=num;
        data[++num].to=x;data[num].next=h[y];h[y]=num;
    }dfs(1,0);build(1,1,n);int mm=read();
    while(mm--){
        int op=read(),x=read();
        if(op==1) add(1,1,n,in[x],out[x],read()%m);
        else{
            node res=ask(1,1,n,in[x],out[x]);int ans=0;
            for(int i=1;i<=tot;++i) if(res.x[prime[i]]) ++ans;
            printf("%d\n",ans);
        }
    }return 0;
}