/*
* LMIO 2019 Tax Evasion
* Main Idea : Euler tour, binary search the answer
* If there's a coin at depth d, it can only climb to
* floor((d - 1) / 2), as Just will get caught afterwards
* So we can build an euler tour, coin X can only stay at
* a range L to R
* Binary search the answer: if a is a candidate answer,
* all coins should have depth at least a - 1
* Check if >= a possible using greedy
*/
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int n, m, timer;
int par[18][200005], depth[200005], lf[200005], rg[200005];
bool is_coin[200005];
vector<int> adj[200005], que[200005], order;
void euler_tour(int nw){
order.push_back(nw);
lf[nw] = timer ++;
for(int nx : adj[nw])
euler_tour(nx);
rg[nw] = timer;
}
int anc(int nw, int dp){
for(int i = 17; i >= 0; i --)
if(dp & (1 << i)) nw = par[i][nw];
return nw;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n >> m;
for(int p, i = 2; i <= n; i ++){
cin >> p;
par[0][i] = p;
depth[i] = depth[p] + 1;
adj[p].push_back(i);
}
for(int i = 1; i < 18; i ++) for(int j = 1; j <= n; j ++)
par[i][j] = par[i - 1][par[i - 1][j]];
for(int p, i = 0; i < m; i ++){
cin >> p;
is_coin[p] = true;
}
euler_tour(1);
for(int i = 1; i <= n; i ++) if(is_coin[i]){
int up = anc(i, (depth[i] - 1) / 2);
que[lf[up]].push_back(rg[up]);
}
int res = 0;
for(int l = 0, r = n, md; l <= r; ){
md = (l + r) / 2;
priority_queue<int, vector<int>, greater<int> > pq;
bool can = true;
for(int i = 0; i < n && can; i ++){
for(int j : que[i]) pq.push(j);
if(depth[order[i]] >= md && !pq.empty()){
int nw = pq.top(); pq.pop();
if(nw <= i) can = false;
}
}
if(!pq.empty()) can = false;
if(can){
res = md; l = md + 1;
}else{
r = md - 1;
}
}
cout << res + 1 << endl;
return 0;
}
