和之前那个 【LNOI】LCA 几乎是同一道题，就是用动态树来维护查分就行. 

code: 

#include <bits/stdc++.h>   
using namespace std;       
#define N 50006  
#define mod 998244353  
#define ll long long 
#define lson t[x].ch[0] 
#define rson t[x].ch[1] 
#define get(x) (t[t[x].f].ch[1]==x) 
#define isrt(x) (!(t[t[x].f].ch[0]==x||t[t[x].f].ch[1]==x))   
#define setIO(s) freopen(s".in","r",stdin) 
int sta[N],hd[N],to[N],nex[N],answer[N],dep[N],n,Q,K,edges;     
inline void add(int u,int v) 
{
    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v;    
}
inline int qpow(int x,int y) 
{
    int tmp=1; 
    for(;y;y>>=1,x=(ll)x*x%mod)   if(y&1) tmp=(ll)tmp*x%mod;  
    return tmp;  
}      
struct sol 
{ 
    int y,id;     
    sol(int y=0,int id=0):y(y),id(id){}   
}; 
vector<sol>a[N];       
struct node 
{  
    int f,rev,ch[2],add;         
    ll sum1,sum2,val1,val2; 
}t[N];            
inline void pushup(int x) 
{   
    t[x].sum1=(t[lson].sum1+t[rson].sum1+t[x].val1)%mod;   
    t[x].sum2=(t[lson].sum2+t[rson].sum2+t[x].val2)%mod;       
}
inline void rotate(int x) 
{
    int old=t[x].f,fold=t[old].f,which=get(x); 
    if(!isrt(old)) t[fold].ch[t[fold].ch[1]==old]=x;   
    t[old].ch[which]=t[x].ch[which^1],t[t[old].ch[which]].f=old; 
    t[x].ch[which^1]=old,t[old].f=x,t[x].f=fold; 
    pushup(old),pushup(x);    
} 
inline void mark(int x,int d) 
{
    (t[x].val2+=1ll*d*t[x].val1%mod)%=mod;     
    (t[x].sum2+=1ll*d*t[x].sum1%mod)%=mod;  
    t[x].add+=d;          
}
inline void pushdown(int x) 
{ 
    if(x&&t[x].add) 
    {
        if(lson)   mark(lson,t[x].add); 
        if(rson)   mark(rson,t[x].add); 
        t[x].add=0;  
    }
}
void splay(int x) 
{ 
    int v=0,u=x,fa; 
    for(sta[++v]=u;!isrt(u);u=t[u].f)     sta[++v]=t[u].f;  
    for(;v;--v)   pushdown(sta[v]); 
    for(u=t[u].f;(fa=t[x].f)!=u;rotate(x))   
    {
        if(t[fa].f!=u) 
        {
            rotate(get(fa)==get(x)?fa:x); 
        }
    }
} 
void Access(int x) 
{
    for(int y=0;x;y=x,x=t[x].f) 
    {
        splay(x); 
        rson=y; 
        pushup(x); 
    }
}         
void dfs(int u) 
{
    dep[u]=dep[t[u].f]+1;   
    t[u].val1=(qpow(dep[u],K)-qpow(dep[u]-1,K)+mod)%mod;    
    for(int i=hd[u];i;i=nex[i])    dfs(to[i]);   
    pushup(u);    
}          
int main() 
{ 
    // setIO("input");  
    int i,j; 
    scanf("%d%d%d",&n,&Q,&K);    
    for(i=2;i<=n;++i) 
    {
        scanf("%d",&t[i].f),add(t[i].f,i);   
    }  
    dep[1]=1,dfs(1);   
    for(i=1;i<=Q;++i) 
    { 
        int x,y; 
        scanf("%d%d",&x,&y);        
        a[x].push_back(sol(y,i));   
    }   
    for(i=1;i<=n;++i) 
    {    
        Access(i),splay(i),mark(i,1);    
        for(j=0;j<a[i].size();++j) 
        {
            Access(a[i][j].y),splay(a[i][j].y);   
            answer[a[i][j].id]=t[a[i][j].y].sum2%mod;   
        }
    }   
    for(i=1;i<=Q;++i)    printf("%d\n",answer[i]);   
    return 0; 
}