離散化後，容易想到設f[i][j]為i節點權值為j的概率，不妨設j權值在左子樹，則有f[i][j]=f[lson][j](p_i·f[rson][1~j]+(1-p_i)·f[rson][j~m])。

　　考慮用線段樹合併優化這個dp。記錄字首和，合併某節點時，若某棵線段樹在該節點處為空，給另一棵線段樹打上乘法標記即可。注意字首和不要算成合並後的了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include
 
<algorithm>
using namespace std;
#define ll long long
#define N 300010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int
 
 x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,p[N],a[N],b[N],fa[N],root[N],t,ans,cnt;
struct data{int to,nxt;
}edge[N];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
 
struct data2{int l,r,x,lazy;
}tree[N<<5];
void ins(int &k,int l,int r,int x)
{
    if (!k) k=++cnt;tree[k].x=1;
    if (l==r) return;
    int mid=l+r>>1;
    if (x<=mid) ins(tree[k].l,l,mid,x);
    else ins(tree[k].r,mid+1,r,x);
}
void update(int k,int x)
{
    if (!k) return;
    tree[k].x=1ll*tree[k].x*x%P;
    if (tree[k].lazy) tree[k].lazy=1ll*tree[k].lazy*x%P;
    else tree[k].lazy=x;
}
void down(int k){update(tree[k].l,tree[k].lazy),update(tree[k].r,tree[k].lazy),tree[k].lazy=0;}
int query(int k,int l,int r,int x)
{
    if (!k) return 0;
    if (l==r) return tree[k].x;
    if (tree[k].lazy) down(k);
    int mid=l+r>>1;
    if (x<=mid) return query(tree[k].l,l,mid,x);
    else return query(tree[k].r,mid+1,r,x);
}
int merge(int x,int y,int l,int r,int s0,int s1,int p)
{
    if (tree[x].lazy) down(x);
    if (tree[y].lazy) down(y);
    if (!x||!y)
    {
        if (!x) x=y,swap(s0,s1);
        update(x,(1ll*p*s1+1ll*(P+1-p)*(P+1-s1))%P);
        return x;
    }
    if (l<r)
    {
        int mid=l+r>>1;
        tree[x].r=merge(tree[x].r,tree[y].r,mid+1,r,(s0+tree[tree[x].l].x)%P,(s1+tree[tree[y].l].x)%P,p);
        tree[x].l=merge(tree[x].l,tree[y].l,l,mid,s0,s1,p),
        tree[x].x=(tree[tree[x].l].x+tree[tree[x].r].x)%P;
    }
    return x;
}
void dfs(int k)
{
    int s=0;
    for (int i=p[k];i;i=edge[i].nxt)
    dfs(edge[i].to),s++;
    if (s==0) ins(root[k],1,m,a[k]);
    else if (s==1) root[k]=root[edge[p[k]].to];
    else root[k]=merge(root[edge[p[k]].to],root[edge[edge[p[k]].nxt].to],1,m,0,0,a[k]);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5461.in","r",stdin);
    freopen("bzoj5461.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++)
    {
        int x=read();
        fa[i]=x,addedge(x,i);
    }
    for (int i=1;i<=n;i++)
    {
        a[i]=read();
        if (p[i]) a[i]=1ll*a[i]*796898467%P;
        else b[++m]=a[i];
    }
    sort(b+1,b+m+1);
    for (int i=1;i<=n;i++) if (!p[i]) a[i]=lower_bound(b+1,b+m+1,a[i])-b;
    dfs(1);
    for (int i=1;i<=m;i++)
    {
        int x=query(root[1],1,m,i);
        ans=(ans+1ll*i*b[i]%P*x%P*x)%P;
    }
    cout<<ans;
    return 0;
}