#include<bits/stdc++.h>
using namespace std;
using pir = pair<int, int>;
const int K = 1320;
const int inf = 1316;
const int N = 1e5 + 5;
pir ans;
vector<pir> s[N];
vector<int> joke, g[N], c[N];
int n, k, id, dd[N], l[N], r[N], ccme[N], sum[N], col[N], dep[N], dfn[N], cnt[N], siz[N];
pir getmx(pir a, pir b){
if(a.first > b.first) return a;
else if(a.first < b.first) return b;
else if(a.second < b.second) return a;
return b;
}
void dfs(int x, int fa){
dep[x] = dep[fa] + 1;
dfn[x] = ++id;
c[col[x]].push_back(x);
for(int v : g[x]){
if(v != fa){
dfs(v, x);
}
}
}
void DFS(int x, int fa, int op){
cnt[dep[x]] += (col[x] == op);
siz[x] += (col[x] == op);
for(int v : g[x]){
if(v != fa){
DFS(v, x, op);
siz[x] += siz[v];
}
}
}
void DDFS(int x, int fa, int op){
s[x].clear();
int maxx = -1, h = 0;
for(int v : g[x]){
if(v != fa){
DDFS(v, x, op);
if(r[v] > maxx){
maxx = r[v], h = v;
}
}
}
if(h) swap(s[x], s[h]), swap(l[x], l[h]), swap(r[x], r[h]), swap(sum[x], sum[h]);
s[x].push_back({dep[x], 1});
sum[x] += min(cnt[dep[x]], 1);
l[x] = dep[x], r[x] = max(r[x], dep[x]);
for(int v : g[x]){
if(v != fa && v != h){
for(pir num : s[v]){
int now = s[x].size() - 1 - (num.first - dep[x]);
int tmp = min(cnt[num.first], s[x][now].second);
s[x][now].second += num.second;
sum[x] += min(cnt[num.first], s[x][now].second) - tmp;
}
}
}
if(col[x] == op) ans = getmx(ans, {sum[x], sum[x] - siz[x]});
}
void DDDFS(int x, int fa){
s[x].clear();
siz[x] = 1;
int maxx = -1, h = 0;
for(int v : g[x]){
if(v != fa){
DDDFS(v, x);
siz[x] += siz[v];
if(r[v] > maxx){
maxx = r[v], h = v;
}
}
}
if(h) swap(s[x], s[h]), swap(l[x], l[h]), swap(r[x], r[h]);
s[x].push_back({dep[x], 1});
l[x] = dep[x], r[x] = max(r[x], dep[x]);
for(int v : g[x]){
if(v != fa && v != h){
for(pir num : s[v]){
int now = s[x].size() - 1 - (num.first - dep[x]);
int tmp = min(cnt[num.first], s[x][now].second);
s[x][now].second += num.second;
}
}
}
if(ccme[col[x]] < inf){
int jian = 0;
int sum = 0;
for(int v : c[col[x]]) dd[dep[v]] = 0;
for(int v : c[col[x]]){
if(dfn[v] >= dfn[x] && dfn[v] <= dfn[x] + siz[x] - 1){
jian++;
}
dd[dep[v]]++;
}
for(int v : c[col[x]]){
if(dd[dep[v]] && dep[v] >= l[x] && dep[v] <= r[x]){
int now = s[x].size() - 1 - (dep[v] - l[x]);
sum += min(dd[dep[v]], s[x][now].second);
dd[dep[v]] = 0;
}
}
ans = getmx(ans, {sum, sum - jian});
}
}
int main(){
ios::sync_with_stdio(0), cin.tie(0);
cin >> n >> k;
for(int i = 1; i <= n; i++){
cin >> col[i];
ccme[col[i]]++;
}
for(int i = 2, f; i <= n; i++){
cin >> f;
g[++f].push_back(i);
}
for(int i = 0; i <= k; i++){
if(ccme[i] >= inf){
joke.push_back(i);
}
}
ans.first = -1;
dfs(1, 0);
for(int i = 1; i <= n; i++) s[i].clear();
for(int v : joke){
for(int i = 1; i <= n; i++) cnt[i] = 0, siz[i] = 0, sum[i] = 0, l[i] = 0, r[i] = 0;
for(int i = 1; i <= n; i++) s[i].clear();
DFS(1, 0, v);
DDFS(1, 0, v);
}
for(int i = 1; i <= n; i++) l[i] = 0, r[i] = 0, siz[i] = 0, s[i].clear();
DDDFS(1, 0);
cout << ans.first << ' ' << ans.second;
return 0;
}
/*
合并子树内有多少个深度为 i 的点
对于每种颜色：
1. 拥有颜色数量 <= sqrt(n)，只需关注需要关注的
2. 大于 sqrt(n)，对于每种颜色单独考虑
*/
