比赛的时候想到是dsu on tree了，但是没想到如何去高效的维护答案，赛后听了大佬的做法后，恍然大悟。然而后面还是调了半天才调处来...看来我对树上启发式合并的理解还不够深刻。
新学了一个卡常小技巧，枚举子树的时候之前是用dfs去遍历的，但是其实可以预处理整棵树的dfs序，之后只要枚举dfs序就可以了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define X first
#define Y second
#define pb push_back
#define pll pair<ll, ll>
#define pli pair<ll, int>
#define pii pair<int,int>
#define New_Time srand((unsigned)time(NULL))
inline ll gcd(ll a, ll b) { while (b != 0) { ll c = a % b; a = b; b = c; }return a < 0 ? -a : a; }
inline ll lowbit(ll x) { return x & (-x); }
int head[2000010], Edge_Num;
struct Edge { int to, next; ll w; }e[4000010];
inline void ade(int x, int y, ll w) { e[++Edge_Num] = { y,head[x],w }; head[x] = Edge_Num; }
inline void G_init(int n) { memset(head, 0, sizeof(int) * (n + 100)); Edge_Num = 0; }
int dir[8][2] = { {-1,0},{0,-1},{-1,-1},{1,-1},{1,0},{0,1},{1,1},{-1,1} };
const long double PI = 3.14159265358979323846;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
inline ll rd() {
    ll x = 0; bool f = 1; char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-')f = 0; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar(); }
    return f ? x : -x;
}
const double eps = 1e-8;
const ll mod = 998244353;
const int M = 1e6 + 10;
const int N = 1e6 + 10;
int fa[N], son[N], siz[N], d[N];
multiset<int>cnt[N];
void dfs(int x) {
    siz[x] = 1;
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].to;
        if (y == fa[x])continue;
        d[y] = d[x] + 1;
        dfs(y);
        siz[x] += siz[y];
        if (!son[x] || siz[son[x]] < siz[y])son[x] = y;
    }
}
int a[N], k, n;
set<int>ext;
inline bool ok(int x, int y) {
    auto it = cnt[x].lower_bound(y);
    if (it == cnt[x].end())return 0;
    return (*it) == y;
}
inline void del(int x, int y) {
    auto it = cnt[x].lower_bound(y);
    cnt[x].erase(it);
}
inline void ins(int x, int y) {
    cnt[x].insert(y);
}
void add(int x, int f) {
    if (f == 1)ins(d[x], a[x]);
    else del(d[x], a[x]);
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].to;
        if (y == fa[x])continue;
        add(y, f);
    }
}
int ans[N];
void gao(int x) {
    if (ok(d[x], a[x] ^ k)) {
        if (ext.find(d[x]) == ext.end()) {
            //++ans[rt];
            ext.insert(d[x]);
        }
    }
    ins(d[x], a[x]);
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].to;
        if (y == fa[x])continue;
        gao(y);
    }
}
void dsu(int x, bool op) {
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].to;
        if (y == fa[x] || y == son[x])continue;
        dsu(y, 0);
    }
    if (son[x]) dsu(son[x], 1);
    for (int i = head[x]; i; i = e[i].next) {
        int y = e[i].to;
        if (y == fa[x] || y == son[x])continue;
        gao(y);
    }
    ans[x] = ext.size();
    ins(d[x], a[x]);
    if (!op)add(x, -1), ext.clear();
}
void solve() {
    n = rd(), k = rd();
    G_init(n);
    for (int i = 2; i <= n; i++) {
        fa[i] = rd(); ade(fa[i], i, 1);
    }
    for (int i = 1; i <= n; i++)a[i] = rd();
    dfs(1);
    dsu(1, 0);
    ll m = 0;
    for (int i = 1; i <= n; i++) {
        m += i ^ (n - ans[i]);
        m %= mod;
    }
    cout << m << endl;
}
int main() {
    int _T = 1;
    //    _T = rd();
    while (_T--)solve();
}