/* 1056. Computer net
* File: Timus
* Author: xiaotian @ hnu
* Created on 2010年10月5日, 下午3:06
* 题解：twice bfs 第一次找离节点1最远的点K，再找离K最远的点X，
* K--X就是最长链，链上的中点就是中心 O(n)
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 10010;
struct node {
int en;
node * next;
node() {
next = NULL;
}
};
node nhead[N];
struct QUEUE {
int val;
int level;
};
QUEUE que[N];
int head, tail;
int line[N];
int flag[N];
int fat[N];
int n;
inline int min(int a,int b) { return a<b?a:b; }
inline int max(int a,int b) { return a>b?a:b; }
void input() {
memset(fat, -1, sizeof (fat));
int en;
for (int i = 2; i <= n; i++) {
scanf("%d", &en);
fat[i] = en;
node *temp1 = (node *) malloc(sizeof (node));
temp1->en = en;
temp1->next = nhead[i].next;
nhead[i].next = temp1;
node *temp2 = (node *) malloc(sizeof (node));
temp2->en = i;
temp2->next = nhead[en].next;
nhead[en].next = temp2;
}
}
void bfs(int sn) {
memset(flag, 0, sizeof (flag));
que[tail].val = sn;
que[tail++].level = 1;
flag[sn] = 1;
int curn;
node *p;
while (head < tail) {
curn = que[head].val;
p = nhead[curn].next;
while (NULL != p) {
if (0 == flag[p->en]) {
flag[p->en] = 1;
que[tail].val = p->en;
que[tail++].level = que[head].level + 1;
}
p = p->next;
}
head++;
}
}
void process() {
head = tail = 0;
bfs(1);
int leaf = que[tail - 1].val;
head = tail = 0;
bfs(leaf);
}
void output() {
line[0] = 0;
int p = tail - 1;
int maxlen = que[tail - 1].level;
line[maxlen] = que[tail - 1].val;
for (int i = maxlen - 1; i >= 1; i--) {
while (que[p].level > i && p >= 0) p--;
while (fat[que[p].val] != line[i + 1] && fat[line[i + 1]] != que[p].val && p >= 0) p--;
line[i] = que[p].val;
}
int x = (maxlen + 1) / 2;
if (1 == (maxlen & 1))
printf("%d\n", line[x]);
else
printf("%d %d\n", min(line[x], line[x + 1]), max(line[x], line[x + 1]));
}
int main() {
while (scanf("%d", &n) != EOF) {
input();
process();
output();
}
return 0;
}